1.00
Introduction
A branch of mathematics where we count number of objects or number of ways of doing a particular job without actually counting them is known as combinatorics and in this chapter we will with elementary combinatorics. For example, if in a room there are five rows of chairs and each row contains, seven hours, then without counting them we can say, total number of chairs is 35. We start this chapter principle of product or fundamental principle of counting.FUNDAMENTAL PRINCIPLE OF COUNTING
1.01
Fundamental Principal Of Counting
MULTIPLICATION PRINCIPLE OF COUNTING If a job can be mm ways, and when it is done in any one of these ways another job be done in n, then both the jobs together can be done in mn ways. The rule can be extended to move number of jobs. Illustration 1.
A college offers 7 courses in the morning and 5 in the evening. Find the possible number of choices with the student if he wants to study one course in the morning and one in the evening.
Solution:
The student has seven choices from the morning courses out of which he can select one course in 7 ways. For the evening course, he has 5 choices out of which he can select one in 5 ways. Hence the total number of ways in which he can make the choice of one course in the morning and one in the evening = 7 x 5 = 35.
Illustration 2.
A person wants to go from station A to station C via station B. There are three routes from A to B and four routes from B to C. In how many ways can he travel from A to C?
Solution:
A B A
Remark:
B in 3 ways C in 4 ways C in3x4 =12ways
The rule of product is applicable only when the number of ways of doing each part is independent of each other i.e. corresponding to any method of doing the first part, the other part can be done by any method
Illustration 3.
How many (1) 5— digit (ii) 3 digit numbers can be formed by using 1, 2, 3, 4, 5 without repetition of digits.
Solution:
(i) Making a 5-digit number is equivalent to filling 5 places. Places:
Number of Choices:
1
2
3
4
5
4
3
2
The first place can be filled in 5 ways using anyone of the given digits.
5 1
The second place can be filled in 4 ways using any of the remaining 4 digits. Similarly, we can fill the 3rd, 4th and 5th place. No. of ways of filling all the five places = 5 x 4 x 3 x 2 x 1 = 120
120
5-digit numbers can be formed. (ii) Making a 3-digit number is equivalent to filling 3 places. Places: l
Number of Choices:
1 5
2
3
4
3
Number of ways of filling all the three places = 5 x 4 x 3 = 60 Hence the total possible 3-digit numbers = 60. Illustration 4.
There are 10 steamers plying between Liverpool and Dublin; in how many ways can a man go from Liverpool to Dublin and return by a different steamer?
Solution:
There are ten ways of making the first passage; and with each of these there is a choice of nine ways of returning (since the man is not to come hack by the same steamer); hence the number of ways of making the two journeys is 10 x 9, or 90. This principle may easily be extended to the case in which there are more than two operations each of which can be performed in a given number of ways.
Addition Principle : If one experiment has n possible outcomes and another has m possible outcomes, then there are (m + n) possible outcomes when exactly one of these experiments is performed. In other words, if a job can be done by n methods and by using the first method, can be done in a1 ways or by second method in 2 ways and so on . . . by the nth method in a ways, then the number of ways to get the job done is (a1 Illustration 5.
a .......... an ) .
A train is going from Cambridge to London stops at nine intermediate stations. Six persons enter the train during the journey with six different tickets. How many different sets of tickets they have had?
Solution:
For S1 ,9 different tickets available, one for each of the remaining 9 stations; similarly at S 2 . 8 different tickets available; and so on.
Hence, it is clear, that total number of different tickets =9+8+7+6+5+4+3+2+1=45 Hence, the six different tickets must be any six of these 45; and there evidently as mar different sets of 6 tickets as there are combinations of 45 things taken 6 at a time. Illustration 6.
A college offers 7 courses in the morning and 5 in the evening. Find the number J of ways a student can select exactly one course, either in the morning or in the evening.
Solution:
The student has seven choices from the morning courses out of which he can select one course in 7 ways. For the evening course, he has 5 choices out of which he can select one course in 5 ways.
Hence he has total number of 7 + 5 = 12 choices. Illustration 7.
A person wants to leave station B. There are three routes from station B o A and J four routes from B to C. In how many ways can he leave the station B.
Solution:
B
A in 3 ways
B
C in 4 ways He can leave station B in 3 + 4 = 7 ways.
1.02
Permutations (Arrangement Of Objects)
The number of permutations of n objects, taken r at a time, is the total number of arrangements of r objects, selected from n objects where the order of the arrangement is important. Without repetition: (a) Arranging n objects, taken r at a time is equivalent to filling r places from n things. r-places 1
No. of choices-
2
n
3
n-1
n-2
4
r
n-3
n-(r-1)
The number of ways of arranging = the number of ways of filling r places =n (n—1) (n—2) (n—r+1) =
n(n 1)(n 2)...................(n r 1((n r )!) (n r )!
n! (n r )!
n
Pr n
(b) The number of arrangements of n different objects taken all at a time
Pn
n!
With repetition: (a) The number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice .... up to r times in any arrangement = The number of ways of filling r places where each place can be filled by any one of n objects. r-places 1
No. of choices-
n
2
n
3
n
4
r
n
n
The number of permutations = the number of ways of filling r places = ( n )
r
(b) The number of arrangements that can be formed using n objects out of which p are identical (and of one kind), q are identical (and of another kind), r are identical (and of another kind) and the rest are distinct is Illustration 8.
How many 7- letter words can be formed using the letters of the words? (a)
BELFAST
(b)
ALABAMA
n! p !q !r !
Solution:
(a) BELFAST has all different letters. Hence the number of words= 7 P7 = 7! = 5040 (b) ALABAMA has 4 A’s but the rest are all different. Hence the number of words that can be formed is 7!/4!= 7 x 6 x 5 = 210.
Illustration 9. a.
How many anagrams can be made by using the letters of the word HINDUSTAN.
b.
How many of these anagrams begin and end with a vowel.
c.
ln how many of these anagrams, all the vowels come together.
d. In how many of these anagrams, none of the vowels come together. e.
In how many of these anagrams, do the vowels and the consonants occupy the same relative positions as in HINDUSTAN.
Solution:
(a) The total number of anagrams = Arrangements of nine letters taken all at a time = 9! /2! = 181440. (b) We have 3 vowels and 6 consonants, in which 2 consonants are alike. The first place can be filled in 3 ways and the last in 2 ways. The rest of the places can be filled in 7! / 2! ways. Hence the total number of anagrams 3 x 2 x 7! / 2! = 15120 (c) Assume the vowels (I, U, A) as a single letter. The letters (IUA) H, D, S, T, N, N can be arranged in 7! / 2! ways. Also IUA can be arranged among themselves in 3! = 6 ways. Hence the total number of anagrams = 7! / 2! x 6 = 15120 (d) Let us divide the task into two parts. In the first, we arrange the 6 consonants as shown below in 6! / 2! ways. x C x C x C x C x C x C x (C stands for consonants and x stands for blank spaces in between them) Now 3 vowels can be placed in 7 places (in between the consonants) in 7 P3 7! — =210 ways. 4! Hence the total number of anagrams =6! / 2! x 210 = 75600. (e) In this case, the vowels can be arranged among themselves in 3! = 6ways. Also, the consonants can be arranged among themselves in 6! / 2! ways. Hence the total number of anagrams 6! / 2! x 6= 2160.
Illustration 10. How many 3 digit numbers can be formed using the digits 0, 1, 2,3,4,5 so that (a) digits may not be repeated (b) digits may be repeated. Solution:
(a) Let the 3-digit number be XYZ Position (X) can be filled by 1, 2,3,4,5 but not 0. So it can be filled in 5 ways.
Position (Y) can be filled in 5 ways again. (Since 0 can be placed in this position). Position (Z) can be filled in 4 ways. Hence by the fundamental principle of counting, total number of ways is 5 x 5 x 4 = l00ways. (b) Let the 3 digit number be XYZ Position (X) can be filled in 5 ways Position (Y) can be filled in 6ways. Position (Z) can be filled in 6 ways. Hence by the fundamental principle of counting, total number of ways is 5 x 6 x 6 = 180
1.03
Important Results
Number of permutations under certain conditions: • Number of permutations of n different things, taken r at a time, when particular thing s to be always included in each arrangement is r n 1 Pr
1
• Number of permutations of n different things, taken r at a time, when a particular thin; is never taken in any arrangement is n 1Pr . • Number of permutations of n different things, taken r at a time, when m particular things are never taken in any arrangement is
n m
Pr
Number of permutations of n different things, taken all at a time, when m specific things always come together is (m!) (n — m + 1)!. • Number of permutations of n different things, taken all at a time, when m specified things never come together is n! — m! (n — m + 1)! Illustration 11. There are m men and n monkeys (n > m). if a man have any number of monkeys .in how many ways may every monkeys have a master? Solution:
The first monkey can select his master by m ways and after that the second monkey can select his master again by m ways, so can the third. And so on, n
hence all monkeys can select master = m x m x m…up to n times = m ways Illustration 12. Find the number of ways in which we can arrange four letters of the word MATHEMAT. Solution:
Letters of the word MATHEMATICS are (M, M), (A, A), (T, T), H, E, I, C and S, making eight distinct letters. We can choose four out of them in 8C4
70 ways, and arrange each of these sets of
four in 4! = 24 ways, yielding (70) (24) = 1680 arrangements. Second, we can choose one pair from
among the three identical letter pairs, and two distinct letters out of the remaining seven in 3
C1
7
C2
3
7 6 /2
63 ways. The letters so obtained can be arranged in 4! /2! = 12
ways, so the number of arrangements in this case is (63) (12) = 756. Finally, we can choose two pairs out of the three identical letter pairs. This can be done in 3C2 ways and the letters obtained can be arranged in 4! /2! 2! = 6 ways, so that the number of arrangements in this last case is (3) (6) = 18. Hence the total number of arrangements is 1680 + 756 + 18 = 2454.
1.04
Cicular Permutations
In the event of the given n things arranged in a circular or even elliptical permutation — and in this case the first and the last thing in the arrangement are indistinguishable — the number of :―permutations is (n—1)!. For example is 20 persons are circularly arranged, the number of arrangements is 19! If positions on a circular arrangement are numbered, then can it be treated as linear arrangement? however, in the case of circular permutations wherein clockwise and anticlockwise orders need not be differentiated — as in the case of differently coloured beads or different flowers made in to a garland, the number of permutations is
n 1! where the number of beads or flowers is taken as n. In details the concept is explained as follows: 2 ARANGEMENTS AROUND A CIRCULAR TABLE Consider five persons A, B, C, D, E be seated on the circumference of a circular table in Order which has no head now, shifting A, B, C, D, E one position in anticlockwise direction we will get arrangements as shown in following figure: We observe that arrangements in all figures are different. Thus, the number of circular permutations of n different things taken all at a time is (n — 1)!, if clockwise and anticlockwise orders are taken as different.
Illustration 13. 20 persons were invited to a party. In how many ways can they and the host be seated at a circular table? In how many of these ways will two particular persons be seated on either side of the host? Solution:
1st part: Total persons on the circular table = 20 guest + 1
= 21 they can be seated in (21 — 1)! = 20! ways.
host
2nd part: After fixing the places of three persons (1 host + 2 persons). Treating (1 host + 2 person) = 1 unit, so we have now ((remaining 18 persons + 1 unit) = 19) and the number of arrangement will be (19 — 1)! = 18! also these two particular person can be seated on either side of the host in 2! ways. Hence, the number of ways of seating 21 persons on the circular table such that two particular persons be seated on either side of the host= 18! x 2! =2x18!
1.05
Arrangements Of Beads Or Flowers (All Different) Around A Circular Necklace Or Garland
Consider five beads A, B, C, D, E in a necklace or five flowers A, B, C, D, E in a garland etc. If the necklace or garland on the left is turned over we obtain the arrangement on the right. i.e., anticlockwise and clockwise order of arrangement is not different we will get arrangements as follows: We see that arrangements in above figures are not different. Then the number of circular permutations of n different things taken all at a time is
1 (n 1)! if clockwise and 2
anticlockwise orders are taken as not different.
Illustration 14. Consider 21 different pearls on a necklace. How many ways can the pearls be placed in on this necklace such that 3 specific pearls always remain together? Solution:
After fixing the places of three pearls. Treating 3 specific pearls = 1 units, so we have now 18 pearls + 1 unit = 19 and the number of arrangement will be (19 — 1)! = 18! also, the number of ways of 3 pearls can be arranged between themselves is 3! 6. Since, there is no distinction between the clockwise and anticlockwise arrangements. So, the required number of arrangements = (18!).
1 18!. 6 = 3 2
NUMBER OF CIRCULAR PERMUTATIONS OF n DIFFERENT THINGS TAKEN r AT A TIME CASE I -
If clockwise and anticlockwise orders are taken as different, then the required number of circular n
permutations=
Pr or r
n
Cr
(r 1)!
Illustration 15. In how many ways can 24 persons be seated round a table, if there are 13 seats? Solution:
In case of circular table the clockwise and anticlockwise order are different, then the required number 24
P13 13
of circular permutations = Case II -
24! 13 11!
If clockwise and anticlockwise orders are taken as not different, then the required number of circular n
permutations =
Pr or 2r
n
Cr
(r 1)! 2
Illustration 16. How many necklace of 12 beads each can be made from 18 beads of various colours? Solution:
In the case of necklace there is not distinction t between the clockwise and anticlockwise arrangements, then the required number of circular permutations 18
P12 2 12
18! 24 6!
18 17 16 15 14 13! 6 5 4 3 2 1 24
119 13! 2
Illustration 17. In how many ways 10 boys and 5 girls can sit around a circular table so that no two girls sit together. Solution:
10 boys can be seated in a circle in 9! ways. There are 10 spaces in between the boys, which can be occupied by 5 girls in
10
P5 ways. Hence total numbers of ways are 9!
10
P5 .or 9! 10 C5 (5!) .
Illustration 18. If n distinct objects are arranged in a circle, show that the number of ways of selecting three of these n things so that no two of them is next to each other is
Solution:
n(n 4)(n 5) 6
Let the n things be x1 , x2 ,..........................xn . The first choice may be one of these n things; and, this is done in n C1 ways. Suppose x1 is the one chosen; the next two may be chosen — excluding x1 and, the two next to x1 , namely x 2 , x n —from the remaining (n —3) in
n 3
C 2 ways. Of these
n 3
C 2 there are (n — 4)
selections when the second two chosen are next to each other, like
x3 x4 .x4 x5 ,..........................xn 2 xn 1 . the number of ways of selecting the second two after x1 is chosen, so that the two are not next to each other is
n 3
C2
(n 3)(n 4) (n 4) 1.2
(n 4)
(n 4)(n 5) 2
The two objects can be relatively interchanged in 2 ways. Further the order of the choice of the three is not to be considered. Hence the number of ways of choice of the three is
n(n 4)(n 5) 2! 2 3!
n(n 4)(n 5) 6
Illustration 19. 2n persons are to be seated n on each side of a long table. r(
Solution:
from (2n — r — s) in
(2 n r s )
Cn
r
ways. Automatically the remaining (n — s) person go to the other
side where already there are s desirous of seating. Thus there are
(2 n r s )
Cn
r
ways of distributing n
persons for each side providing for the restriction of on one side and s on the other side. n persons on each side can be permuted in n seats in n! ways. The number of ways of seating the 2n persons, n on each side, is therefore (2 n r s )
Cn
2
r
n! S
Combinations
1.06
Each of different grouping or selections that can be made by some or all of a number of given things without considering the order in which things are placed in each group, is called COMBINATIONS
Can combinations be considered for both non-identical and identical objects?
COUNTING FORMULAE FOR COMBINATIONS The number of combinations (selections or groupings) that can be formed from n different objects taken r at a time is denoted by n C r and its value is equal to n
n! 0 n r !r !
Cr
n
r
n as n Cr
Pr ! r!
as in a permutation the arrangement of r selected objects out of n, is done in r! ways and in combination arrangement in a group is not considered. In particular
n
C0
n
Cn
1 i.e. there is only one way to select none or to select all objects out of n distinct objects.
n
C1
n There are n ways to select one thing out of n distinct things.
n
Cr
n
Cn
n
r
therefore Cx
n
Cy
x
y or x y n
If n is odd then the greatest value of n C r is Cn
1
n
or n Cn
2
1
2
n
If n is even then the greatest value of n C r is Cn 2
If the number of ways of selection of r objects out of 2n objects is maximum then, What can be the possible value of r?
Illustration 20. Six X’s have to be placed in spaces on the adjoining Figure so that each row contains at least one X. In how many different ways this can be done?
Solution:
If there be no restriction on the placing of the) (s, the number of ways is 8 C6
8
C2
28 these
there are two ways in which the X; s can be placed; one, with the first row empty and the other with the third row empty. These two cases only do not satisfy the condition. So the number of ways = 28— 2 = 26. IMPORTANT RESULTS OF COMBINATIONS (SELECTIONS) The number of ways in which r objects can be selected from n distinct objects if a particular object is always included is
n 1
Cr 1 .
The number of ways in which r objects can be selected from n distinct objects if a particular object is always excluded is
n 1
Cr .
The number of ways in which r objects can be selected from n distinct objects if m particular objects are always included is n n m Cr
m
.
The number of ways in which r objects can be selected from n distinct objects if m particular objects are always excluded is n n m Cr . Illustration 21. A lady desires to give a dinner party for 8 guests. In how many ways can the lady select guests for the dinner from her 12 friends, if two of the guests will not attend the party together? Solution:
The following three methods of approach are indicated. (i) Number of ways of forming the party
=
12
C8
10
C6 since
10
C6 is the number of ways of making up the party with both the specified
guests included. =495—210=285 (OR) (ii) Number of ways of forming the party = Number of ways of forming without both of them + Number of ways of forming with one of them and without the other =
10
2.10 C7
C8
45 240
285 (OR)
(iii) Split the number of ways of forming the party = those with one of the two (say A) + those without A = Note-
10
2.11 C8
C7
120 165
285
The number of ways in which r objects can be selected from n objects if m particular r
r n m
objects are identical is
n m
Cr or
r 0
Cr according as r
m or r
m
r r m
Illustration 22. A bag contains 23 balls in which 7 are identical. Then find the number of ways of selecting 12 balls from bag. Solution:
Here n=23, p=7, r=12(r>p) 12 16
Hence, required number of selections
Cr
r 5
16
C5 16
17 17
C6 C11 17
18
1.07
C5
C11
C12
16
C6
16 17
C9
17 18
16 17 17
C8
16
C7
16
C8
C10 C10 17
C12
C10
16
C7
C6
C8
17
16
18
C6
17
C9
16 16
16
C10
C9
16
C11
16
C10
C12
16
C11
C12
n Cr
n
C12
n Cr
n
17
C9 18
17
C10
C8
Selection From Distinct / Identical Objects
SELECTION FROM DISTINCT OBJECTS
16
C12 n 1
Cr
1
Cn
r
Cr
The number of ways (or combinations) of selection from n distinct objects, taken at least one of them is n
C1
n
C2
n
C3
..........................................
n
Cn
2n
1
Logically it can be explained in two ways, as one can be selected in n C1 ways, two in n C 2 ways and so on and by addition principle of counting the total number of ways of doing either of the job is n
C1
n
C2
n
C3
..........
n
Cn
Also, for every object, there are two choices, either selection or non-selection. Hence total choices are 2 n . But this also includes the case when none of them is selected. Therefore the number of selections, when at least one is selected =
2n 1
It is allowed to select at most 5 things out of 11 different things. In how many ways at least one of them can be selected? Illustration 23. Given five different green dyes, four different blue dyes and three different red dyes, how many combination of dyes can be chosen taking at least one green, one blue dye? Solution:
Any one dye of a particular colour can be either chosen or not; and, thus there are 2 ways in which each one may be dealt with. Number of ways of selection so that at least one green dye is included
25 1 31
(1 is subtracted to correspond to the case when none of the green dyes is chosen.) A similar argument may be advanced in respect of other two colours also. Number of combinations = 25
1 24 1 23
31 15 8 3720
SEL ECTION FROM IDENTICAL OBJECTS 1. The number of selections of r r
n objects out of n identical objects is 1.
2. The number of ways of selections of at least one object out of n identical object is n. 3. The number of ways of selections of at least one out of a1 , a2 , a3 ...........an objects, where a1 are alike of one kind, a 2 are alike of second kind, and so on a n are alike of the nth kind, is
a1 1 a2 1 ................... an 1
1
4. The number of ways of selections of at least one out of a1 alike of one kind a are alike of nth kind and k are distinct is
a1 1 a2 1 ................... an 1 2k 1
a2
a3 ................... an
k objects, where a1 are
Illustration 24. Find the number of combinations that can be formed with 5 oranges, 4 mangoes and 3 bananas when it is essential to take (i) Solution:
at least one fruit
(ii) one fruit of each kind.
Here 5 oranges are alike of one kind, 4 mangoes are alike of second kind and 3 bananas are alike of third kind (i) The required number of combinations (when at least one fruit) = (5 + 1) (4 + 1) (3 + 1)2°—i = 120—1 =119 (ii) The required number of combinations (when one fruit of each kind) 5
1
,
2
N and n ,
3
C1
C1
5 4 3
60
Divisors Of A Given Natural Number
1.08 Let n
4
C1
3
P1 1 .P2 2 .P3 3 .....................Pk k , where P1 , P2 , P3 ..............Pk are different prime numbers and
..............
are natural numbers then:
k
the total number of divisors of N including 1 and n is 1
1,
2 ,
2
3
3 ..............
1
k
the total number of divisors of n excluding 1 and n is 1
1,
2 ,
2
3
3 ..............
1
k
2
the total number of divisors of n excluding exactly one out of 1 or n is 1
1,
2 ,
2
3
3 ..............
the sum of these divisors is = P01 , P11 , P12
.......P1 1
1 2
1
1
k
1,
2
1
1 ,..............
P02 , P21 , P22
.......P2
2
k
1
.......... P0 k , Pk1 , Pk 2 .......Pk
k
(Use sum of G.P. in each bracket) the number of ways in which n can be resolved as a product of two factors is if n is not a perfect square
1 2
1
1,
2
1 ,..............
k
1
1 if n is a perfect square
the number of ways in which is composite number n can be resolved into two factors which are relatively prime (or co prime) to each other is equal to 2
k 1
where k is the number of different factors (or different primes) in n.
Illustration 25. If n = 10800then find the (a)
total number of divisors of n
(b) the number of even divisors
(c)
the number of divisors of the form 4m + 2
(d) the number of divisors which are multiples of 15 n= 10800 = 2
Solution:
4
33 52
Any divisor of n will be of the form 2
a
3b 5c where 0
a
4,0
b
3,0
c
2.
For any distinct choices of a, b and c, we get a divisor of n (a) total number of divisors = (4 + 1) (3 + 1) (2 + 1) = 60 (b) for a divisor to be even, a should be at least one. So total number of even divisors = 4(3 + 1) (2 + 1) = 48. (c) 4m + 2 = 2(2m + 1). In any divisor of the form 4m + 2, a should be exactly 1. So number of divisors of the form 4m + 2 = 1(3 + 1) (2 + 1) = 12. (d) A divisor of n will be a multiple of 15 if b is at least one and c is at least one. So number of such divisors = (4+1) x 3 x 2 = 30. (8) Division into groups: (i) The number of ways in which (m + n) different things can be divided into two groups which contain m and n things respectively is m n
Cm nCn
(m n)! ;m n m!n !
Corollary : If m = n, then the groups are equal size. Division of these groups can be given by two types. Type I : If order of group is not important: The number of ways in which 2n different things can be divided equally into two groups is
(2n)! 2!( n !) 2 Type II: If order of group is important: The number of ways in which 2n different things can be divided equally into two distinct groups is
(2n)! 2n ! 2! 2 2!(n !) ( n !) 2 (ii) The number of ways in which (m + n +p) different things can be divided into three groups which contain m , n and p things respectively is m n p
Cm n p C n p C p
(m n p )! ;m m !n ! p !
n
p
Corollary : If m = n = p, then the groups are equal size. Division of these groups can he given by two types. Type .I : If order of group is not important: The number of ways in which 3p different things can be divided equally into three groups is
(3 p )! 3!( p !)3
Type II : If order of group is important : The number of ways in which 3p different things can be divided equally into three distinct groups is
(3 p)! 3 p! 3! 3 3!( p !) ( p !)3 Note: (1) If order of group is not important: The number of ways in which mn different things can be divided equally into m groups is
mn ! ( n !) m m !
(2) If order of group is important : The number of ways in which mn different things can be divided equally into m distinct groups is
mn ! (mn)! m! m (n !) m ! (n !) m
Illustration 26. In how many ways can a pack of 52 cards be equally among 4 players in order? . ‘ Solution.
Here order ot group is important, then. the numbers of ways in which 52 different cards can be divided equally into 4 players is
52! 4!(13!)4
4!
(52)! (13!) 4
Alternative method : Each player will get 13 cards. Now first player can be given 13 cards out of 52 cards in 52C13 ways. Second player can be given 13 cards out of remaining 39 cards (i.e. 52— 13 = 39) in 39C13 ways. Third player can be given 13 cards out of remaining 26 cards (i.e., 39 — 13 = 26) in 26C13 ways and fourth player can be given 13 cards out of remaining 13 cards (i.e. 26— 13 = 13) in 13C13 ways. Hence required number of ways 52
39
C13
C13
26
C13
13
C13
52! 39! 26! 1 13! 39! 13! 26! 13! 13! 52! 13!
4
Illustration 27. In how many ways can a pack of 52 cards be formed into 4 groups of 13 cards each ? Solution.
Here order of group is not important, then the number of ways in which 52 different cards can be divided equally into 4 groups is
52! 4!(13!) 4 Alternative Method: Each group will get 13 cards. Now first group can be given 13 cards out of 52 cards in C13 ways. Second group can be given 13 cards out of remaining 39 cards (i.e. 52— 13 = 39) in 39C13 ways. Third group can be given 13 cards out of remaining 26 cards (i.e., 39 — 13 = 26) in 26C13 ways and fourth group can be given 13 cards out of remaining 13
cards (i.e.,26 — 13 = 13) in 13C13 ways. But the all (four) groups can be interchanged in 4 ! ways. Hence the required number of ways 1 4! 52! 39! 26! 1 1 13! 39! 13! 26! 13! 13! 4! 52!
52
39
C13
C13
26
C13
13
C13
4
13! 4!
Illustration28. In how many ways can a pack of 52 cards be divided in 4 sets, three of them having 17 cards each and fourth just card? Solution.
First we divide 52 cards into two sets which contains 1 and 51 cards respectively is
(52)! 1!51!
Now 51 cards can be divided equally in three’ sets each contains 17 cards (Here order of sets is not important) in
(51)! ways. 3!(17!)3
Hence the required number of ways
(52)! (51)! 1!51! 3!(17!)3 (52)! 1!3!(17!)3
(52)! (17!)3 3!
Alternative Method : First set can be given 17 cards out of 52 cards in 52C17. Second set can be given 17 cards out of remaining 35 cards (i.e. 52 — 17 = 35) in 35C17. Third set can be given 17 cards out of remaining 18 cards (i.e., 35 — 17 = 18) in 18C17 and fourth set can be given 1 card out of 1 card in 1C1. But the first three sets can be interchanged in 3 ! ways. Hence the total number of ways for the required distribution 1 3! 52 35! 18! 1 1 17! 35! 17! 18! 17! 1! 3! (52)! (17!)3 3!
52
C17
35
C17
18
C17
1
C1
Illustration29. In how many ways can 12 balls be divided between 2 boys, one receiving 5 and the other 7 balls? Also in how many ways can these 12 balls be divided into groups of 5, 4 and 3 balls respectively ? Solution.
I Part: Here order is important, then the number of ways in which 12 different balls can be divided between two boys which contains 5 and 7 balls respectively, is
(12)! 2! 1584 5!7!
Alternative Method: First boy can be given 5 balls out of 12 balls in 12C5. Second boy can be given 7 balls out of 7 balls (i.e. 12 — 5 = 7) but there order is important (boys interchange by (2 types) then required no. of ways 12
C5
7
C7
2!
12! 1 2! 5! 7! 12! 2 5! 7! 12. 11. 10. 9. 8. 7! .2 5. 4. 3. 2. 1. 7! 1584 II part : Here order is not important then the number of ways in which 12 different balls can be
divided into three groups of 5, 4 and 3 balls respectively, is
(12)! 27720 5!4!3! Alternative Method: First group can be given 5 balls out of 12 balls in C5 ways. Second group can be given 4 balls out of remaining 7 balls (12 — 5 = 7) in 7C4 and 3 balls can be given out of remaining 3 balls in 3C3. Hence the required number of ways (Here order of groups are not important) 12
C5
7
C4
3
C3
12! 7! 1 5! 7! 4! 3! 12! 5! 4! 3!
(9) Arrangement in Groups: (i) The number of ways in which n different things can be arranged into r different groups is n r 1
Pn or n !n 1 Cr
1
according as blank group are or are not admissible. Illustration 30. In how many ways 5 different. balls can be arranged into 3 different boxes so that no box remains empty? Solution.
The required number of ways = 5!.5 1 C3
1
5!.4 C2
720
Alternative Method : Each box must contain a least one ball since no box remains empty. Boxes can have balls in the following systems
All 5 balls can be arranged by 5 ! ways and boxes can be arranged in each system by
Hence required number of ways 5!
3! 2!
3! 3! 5! 720 2! 2!
(ii) The number of ways in which n different things can be distributed into r different groups is
rn
r
C1 (r 1) n
r
2) n .................... ( 1) r
C2 ( r
coefficient of xn in n !(e
x
1r
Cr
1
1) r
Here blank groups are not allowed. Note Coefficient of xr in e px
pr r!
Illustration 31. In how many ways 5 different balls can be distributed into 3 boxes so that no box remains empty? Solution.
The required number of ways 35 – 3 C1 3 – 1
5
3
5
C2 3 – 2 – 3 C3 3 – 3
5
243 – 96 3 – 0 150 OR Coefficient of x in 5! (e x – 1) 3 5
coeffi of x 5 in 5! e3 x – 3e 2 x 5!
35 25 –3 5! 5
3
3e x – 1
1 5!
35 – 3.25 3 243 – 96 3 150
Alternative Method : Each box must contain at least one ball since no. box remains empty. Boxes can have balls in the following systems
The number of ways to distribute the balls in I system = 5 C1
4
The total number of ways to distribute 1, 1, 3 balls to the boxes 5
= C1
4
C1
3
C3
3! 2!
5 4 1 3 60 and the number of ways to distribute the balls in II system = 5 C1
4
C2
2
C2
C1
3
C3
The total number of ways to distribute 1, 2, 2 balls to the boxes 5
= C1
4
2
C2
3! 2!
C2
5 6 1 3 90 The required number of ways = 60 + 90 = 150 (iii) The number of ways in which n identical things can be distributed into r different groups is n r 1
Cr
1
or
n 1
Cr
1
according as blank groups are or are not admissible. Illustration 32. In how many ways 5 identical balls can be distributed into 3 different boxes so that no box remains empty? Solution.
The required number of ways =
5 1
C3
1
= 4C2 = 6
Alternative Method: Each box must contain at least one ball since no box remains empty. Boxes can have balls in the following systems
Here balls are identical but boxes are different the number of combinations will be I in each systems. Required number of ways =1
3! 3! 1 6 2! 2!
Illustration 33. Four boys picked up 30 mangoes. in how many ways can they divide them if all mangoes be identical? Solution.
Clearly, 30 mangoes can be distributed among 4 boys such that each boy can receive any number of mangoes. Hence total number of ways 30 4 1
C4
1
= 33C3 =5456
Illustration 34. Find the positive number of solutions of x + y + z + w = 20 under the following conditions:
Solution.
(i)
zero values of x, y , z, w are included.
(i)
Since x + y + z + w = 20
Here x
(ii) zero values are excluded.
0, y 0, z 0, w 0
The number of solutions of the given equation in this case is same as the number of ways of distributing 20 things among 4 different groups. Hence total number of solutions = 20 (ii) Since x+y+z+w = 20
4 1
C4
23 1
C3
1771
Here
x 1, y 1, z 1, w 1
or
x –1 0, y –1 0, z –1 0, w –1 0
Let
x1
x –1
x
x1 1
y1
y –1
y
y1 1
z1
z –1
z
z1 1
w1
w –1
w1 1
w
Then from (1) x1 1 y1 1 z1 1 w1 1 20
and
x1
y1
z1
x1
0, y1
w1
16
0, z1
0, w1
0
16 4–1
Hence total number of solutions
C4–1
19. 18. 17 1. 2. 3 57. 17 969. 19
C3
Alternative Method :
Part (ii ) :
x
y z w
20
x 1, y 1, z 1, w 1 20–1
Hence total no. of solutions
C4–1
19
C3
969.
Illustration 35. How many integral solutions are there to x + y + z + t = 29, when x 1,y l, z 3andt 0? Solution.
Since x + y + z + t = 29 and x , y, z, t are integers x 1, y
2, z
3, t
x – 1 0, y – 2 let or x
x – 1, x2
x1 x1 1, y
x2
2, z
0
0, z – 3 0, t y – 2, x3
0
z –3
x3 3 andthen x1
0, x2
0, x3
0, t
0
From (1), x1 1 x2 x1
x2
2 x3 3 t x3 t
23
Hence total number of solutions 26
C3
29
26. 25. 24 1. 2. 3
23 4–1
C4–1
2600
Illustration 36. How many integral solutions are there to the system ‘of equations x1 +x2 +x3 +x4+x5 = 20 and x1 +x2 = 15 when xk Solution.
0?
We have x1 +x2 +x3 +x4+x5 = 20 and x1 + x2 = 15 then from (i) and (ii) we get two equation
………………(i) ………………(ii)
x3
x4
x5
x1
x2
15
5
and given x1
0, x2
0, x3
0, x4
0 and x5
0 5 3–1
Then number of solutions of equation (iii) 7
7. 6 1. 2
C2
C3–1
21
and number of solutions of equation (iv) = 15 + 2— 1C2-1 = 16C1 = 16 Hence total number of solutions of the given system of equations = 21 x 16 =336. Illustration 37. Find the number of non-negative integral solutions of 3x + y + z = 24. Solution.
We have let x = k 3x + y + z = 24 and given x
0, y
0, z
0
y+z = 24—3k Here
24
Hence 0
k
24 – 3k
0
8
The total number of integral solutions of (i )is 24–3 k 2–1
C2–1
25–3 k
C1
25 – 3k Hence the total number of solutions of the original equation 8
8
25 – 3k k 0
25. 9 – 3.
(iv)
8.9 2
225 – 108
25
8
1– 3 k 0
k k 0
117
The number of ways in which n identical things can be distributed into r groups so that no group contains less than l things and more than in things (l < m) is ii. 1 1+1 1+2 in r coefficient of xn in the expansion of x
l
or coefficient of xn in the expansion of x
xl lr
1
1- x m
xl l 1 r
2
1 x
xm
...
r
r
Illustration 38. In how many ways can three persons, each throwing a single dia once, make a sum of 15? Solution.
Numbers on the faces of the dia are 1, 2, 3, 4, 5, 6 (least number 1, greatest number 6)
coefficient of x15 in x1
Required number of ways
coeffi. of x15 in x 3 1 x x 2 coeffi. of x12 in 1 x x 2 3
coeffi. of x12 in 1 – x 6
x3 x3
1– x
x4 x4
x3
x4
x5
x6
3
3
x x5
x2
3
–3
coeffi. of x12 in 1 – 3 x 6 3 x12 1 3 x 6 x 2 .... 28 x 6 ... 91x12 ... 91 – 84 3
94 – 84
10
Illustration 39. In how many ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question. Solution.
If examiner given marks any seven question 2 (each) marks then marks on remaining questions given by examiner = — 7 x 2+ 30 = 16 If x are the marks assigned to ith question, then x1 + x2 + x3 + ... + x8 = 30 and 2
xi
l6fori =1,2,3
8. Therefore the required number of ways 30
coefficient of coeffi. of
30
in
coeffi. of
14
in 1
coeffi. of
14
in
coeffi. of
14
in 1 –
coeffi. of
14
in 1
21
2
in 16
3
1
14 8
.... 14 8
...
1– 1–
15
16 8
...
8
–8
8
C1
9
C2
2
...
21
C13
14
....
21
C14
C7
Note: Coefficient Of xr in the expansion of(l —x)-n is n + r - 1Cr (v) If a group has n things in which p are identical, then the number of ways of selecting r things from a group is r r 0
n– p
r
Cr or r r– p
n– p
Cr according as r
p or r
p.
Illustration 40. A bag has contains 23 balls in which 7 are identical. Then find the number of ways of selecting 12 balls from bag. Solution.
Here n = 23,p=7,r=12 (r > p) 12 16
Required number of selections
Cr
r 5 16
16
C5
16
C6
17
C11
17
16
C5
17
C6
17
17
18
C12
18
C6
C9
18 18
16
C7
C6
C8
17
C11
16
17
17
16
C7
C10
17
C10 17
C12
16
C8 16
C9
16 16
C8
C12 n Cr
17
C9
16
C9 n
C12 n Cr 17
16
C10
16
C12
16
C10
Cr –1
n
C11
n 2
C11
16
C12
Cr
Cn – r
C10
C10 C8
Deragements If. n things are arranged in a row, the number of ways in which they can be deranged so that no one of them occupies its original place is
n! 1
1 1 1 1 1 ....................... ( 1) n 1! 2! 3! 4! n!
No object goes to its scheduled place. Remark: If r things goes to wrong place out of n things then (n — r) things goes to original place (Here r < n) If Dn = No. of ways, if all n things goes to wrong place and Dr = No. of ways, if r things goes to wrong place If r goes to wrong places out of n, then (n — r) goes to correct places. Then Dn = nCn - r Dr Where Dr
r! 1
1 1 1 1 1 ....................... ( 1) r 1! 2! 3! 4! r!
Illustration 41. A person writes letters to six friends and addresses the corresponding envelopes. in how many ways can the letters he placed in the envelopes so that (i) at least two of them are in the wrong envelopes. (ii) all the letters are in the wrong envelopes. 6
Solution.
n
(i ) The number of ways in which at least two of them in the wrong envelopes r 2 n
Here n
Cn – 2 D2
n
Cn –3 D3
n
Cn – 4 D4
n
Cn –5 D5
n
Cn – 6 D6
6 6
6
C4 .2! 1 –
C1 .5! 1 –
1 1 1! 2!
6
C3 .3! 1 –
1 1 1 1 1 – – 1! 2! 3! 4! 5!
1 1 1 – 1! 2! 3! 6
C0 6! 1 –
6
C2 .4! 1 –
1 1 1 1 – 1! 2! 3! 4!
1 1 1 1 1 1 – – 1! 2! 3! 4! 5! 6!
Cn – r Dr
15 40 135 264 265 719 (ii ) The number of ways is which all letters be placed in wrong envelopes 6! 1 – 720
1 1 1 1 1 1 – – 1! 2! 3! 4! 5! 6!
1 1 – 2 6
1 1 –120 24 720
360 –120 30 – 6 1 265
Alternative Method: (i) The number of all the possible ways of putting 6 letters into 6 envelopes is 6 !. Number of ways to place all letters correctly into corresponding envelopes = 1 and Number of ways to place one letter is the wrong envelope and other 5 letters in the write envelope = 0 (: It is not possible that only one letter goes in the wrong envelop when if 5 letters goes in the right envelope, then remaining one letter also goes in the write envelope) Hence number of ways to place at least two letters goes in the wrong envelopes =6!—0—l =6! —1 = 720 —1 =719 (ii) The number of ways I letter in 1 address envelope so that one letter is in wrong envelope=0 (Because it is not possible that only one letter goes in the wrong envelope) The number of ways to put 2 letters in 2 addressed envelopes so that all are in wrong envelopes = The number of ways without restriction — The number of ways in which all are in correct envelopes — The number of ways in which 1 letter is in the correct envelop =2!—l—0 =2—1 =1 (2){from(1)} The number of ways to put 3 letters in 3 addressed envelopes so that all are in wrong envelopes = The number of ways without restriction — The number of ways in which all are in correct envelopes — The number of ways in which 1 letter is in the correct envelope — The number of ways in which 2 letter are in correct envelope 3!—l—3C1 x l—0 (from(1),(2)} =2 (3C1 means that select one envelope to put the letter correctly) The number of ways to put 4 letters in 4 addressed envelopes so that all are in wrong envelopes The number of ways without restriction — The number of ways in which all are in correct envelopes — The number of ways in which I letter is in the correct envelope — The number of ways in which 2 letter are in correct envelopes — The number of ways in which 3 letters are in correct envelopes
4! –1 – 4 C1 2 – 4 C2 1 – 4 C3 0 24 –1 – 8 – 6 9
The number of ways to put 5 letters in 5 addressed envelopes so that all are in wrong envelopes = The no. of ways without restriction — The no. of ways in which all are in correct envelopes — The no. of ways in which 1 letter is in correct envelop — The no. of ways in which 2 letters are in correct envelopes — The-no, of ways in which 3 letters are in correct envelopes — The no. of ways in which 4 letters are in correct envelopes 5! –1 – 5C1 9 – 5C2 2 – 5C3 1 – 5C4 0 120 –1 – 45 – 20 –10 – 0 44
The number of ways-to put 6 letters in 6 addressed envelopes so that all are in wrong envelopes = The no. of ways without restriction — The no. of ways in which all are in correct envelopes — The no. of ways in which I letter is in the correct envelop — The no. of ways in which 2 letters are in correct envelopes — The no. of ways in which 3 letters are in correct envelopes — The no. of ways in which 4 letters are in correct envelopes — The no. of ways in which 5 letters are in correct envelopes. 6! –1 – 6 C1 44 – 6 C2 9 – 6 C3 2 – 6 C4 1 – 6 C5 0 {from (1), (2), (3), (4) & (5)} 720 –1 – 264 –135 – 40 –15 720 – 455
265.
Multinomial Theorem (i) If there are 1 objects of one kind, m objects of second kind, n objects of third kind and so on; then the number of ways of choosing r objects out of these objects (i.e.,l + m + n + ...) is the coefficient of xr in the expansion of (l+x+x2+...+xl(l+x+x2+...+xm)( l+x+x2+...+xn) Further if one object of each kind is to be included, there the number of ways of choosing r objects out of these objects (i.e., 1 + m + n + ...) is the coefficient of x in the expansion of (x + x2 ± x3 + ... x’) (x + + + ... X) (x + + x3 + ... + x) (ii) If there are I objects of one kind, m object of second kind, n object of third kind and so on; then the number of possible arrangements/permutations of r objects out of these objects (i.e. 1+ m + n + ...) is the coefficient of x in the expansion of
r! 1
x x2 xl ............ 1! 2! l!
1
x x2 xm x x2 xn .......... 1 ........... 1! 2! m! 1! 2! n!
Illustration 42. Find the number of Combinations and Permutations of 4 letters taken from the word EXAMINATION. Solution.
There are 11 letters A, A; I, I; N, N; E, X, M, T, O. Then No. of combinations
coefficient of x 4 in 1 x x 2
3
1 x
5
2 A ' s, 2 I ' s, 2 N ' s,1E ,1X ,1M ,1Tand1O coefficient of x 4 in
1 x
coefficient of x 4 in
1 x
8
C4
0 3.7 C2
3
3
x6 3 1 x
8
x6 1 x
5
2
x2 3 1 x x4 1 x
3x 2 1 x
7
3x 4 1 x
5
6
and No. of Permutations x coefficient of x in 4! 1 1!
x2 2!
coefficient of x 4 in 4! 1 x
x2 2
4
coefficient of x 4 in 4!
1 x
coefficient of x 4 in 4!
1 x
4! 24
0
37 . C2 2
3 4
8.7.6.5 1.2.3.4
3 7.6 . 2 1.2
3 4
8
C4
8. 7. 6. 5 6 3. 7. 6
6. 3
3
8
3
5
x 1 1!
3
1 x x6 8
5
3 1 x 2
x6 1 x 8
5
2
x2
3 4 x 1 x 4
3 2 x 1 x 2
7
1 x
5
3 4 x 1 x 4
6
1680 756 18
Alternative Method : There are 11 letters: A, A;I, I;N, N;E, X, M, T, O. The following cases arise: Case I : All letters different : The required number of choosing 4 different letters from 8 different (A, I, N, E, X, M, T, 0) types of the letters = 2454 8
C4
8.7.5.6 1.2.3.4
70
8 and No. of Permutations = 8P4 = 8.7.6.5 = 1680 Case II : Two alike of one type & Two alike of another type 2A’s, 2I’s, or 2I’s, 2N’s, or 2N’s, 2A’s.: This must be
= 18 Case III: Two alike and Two different: This must be 2A’s or 21’s or 2N’s and for each case 7 different letters.
= 756 From Case I,II and III, The required No. of Combinations =70+3+63 and No. of Permutations = 1680 + 18 + 756 =2454. How to find number of Solutions of the Equation If the equation 2
3
... q
n
.....(1)
(i ) If zero included then no. of solutions of (1) coefficient of x n in 1 x x 2 ... 1 x 2
1 x3
x 6 ... ... 1 x q
x 4 ...
x 2 q ... –1
coefficient of x n in 1 – x
–1
1 – x2
1 – x3
–1
–1
... 1 – x q
(ii ) If zero exluded then the no. of solutions of (1) coefficient of x n in x x 2
x 3 .... x 2
x4
x 6 ... x 3
x6
x 9 ...
.... x q coefficient of x n in x1 coefficient of x
n–
2 3 ... q
q q 1 2
–1
1– x
in 1 – x
–1
1– x
2 –1
–1
1 – x2
1– x –1
1 – x3
3 –1
... 1 – x
x2q
–1
... 1 – x q
Illustration 43. Find the number of non negative integral solutions of x1 x2 Solution.
...
q –1
x3
4 x4
20 .
Number of non negative integral solutions of the given equation –1
coefficient of x 20 in 1 – x
1– x
–1
1– x
18
C16 x16 ...
22
1 – x4
–1
–1
–3
coefficient of x 20 in 1 – x 1 – x4 20 3 coefficient of x in (1 C1 x 4 C2 x 2
....
–1
5
C3 x 3
6
C4 x 4 ...
10
6
1
6
C4
10
C2
10
6. 5 1. 2
1
C8
14
C2
14
...
x8
x12
14
C12 x12 ..
C20 x 20 ...) 1 x4
1
C8 x8
C12 C2
10. 9 1. 2
18 18
22
C16
22
C2
x16
x 20 ...
C20
C2
14. 13 1. 2
18. 17 1. 2
22. 21 1. 2
1 15 45 91 153 231 536
Illustration 44. Find the number of positive unequal integral solution of the equation x + y + z + w = 20; Solution.
We have x
y z w
20
assume x
y
now let x
x1 , y – x
x
w, Here x, y, z , w 1 x2 , z – y
x3 and w – z
x2 , z
x1
x2
from (1), 4 x1 3 x2
2 x3
x4
20, then x1 , x2 , x3 , x4
x4
20
4 x1 3 x2
2 x3
x3 , w
x1
x4
x1
x1 , y
z
x2
x3
x4 1 ......(2)
Number of positive integral solutions of (2) coefficient of x 20–10 in 1 – x 4
–1
1 – x3
–1
1 – x2
–1
1– x
–1
x6
x8
x10
x9
x4
x6
x8
x10
x7
x9
x10
x3
x4
x5
x6
x7
x8
x9
x10 )
x8
x10 )(1 x x 2
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 23
Number of Rectangles and Squares n
(i)
n
r 3 and number of squares of any size is
Number of rectangles of any size in a square of n x n is r 1
r2 r 1
(ii) In a rectangle of n x p (n
np (n 1)( p 1) and number of squares of any 4
n
(n 1 r )( p 1 r )
size is r 1
Illustration 45. Find the number of rectangles excluding squares from a rectangle of size 9 × 6. Solution.
Here n =6 and p =9. No. of rectangles excluding square 6 6.9 (6 1)(9 1) – (7 – r )(10 – r ) 4 r 1 6
945 –
(70 –17r r 2 )
r 1
945 –154
791.
Exponent of Prime p inn! Since exponent of prime p in n is denoted by Ep (n !). where p is a prime number and n is a natural number. Then the last integer amongst 1, 2, 3, ...,(n — 1), n which is divisible by p is
n n p, where denotes the greatest integer. p p
[ x] E p (n !)
x
E p (1.2.3.....(n –1).n) n p p
E p p.2 p.3 p.....(n –1) p.
because the remaining natural numbers from 1 to n are not divisible by p. n p
E p 1.2.3....
n p n which is divisible by p is p
Now the last integer amongs 1, 2,3...., n. p p
n p2
n p
E p p.2 p.3 p....
n p p2
because the remaining natural numbers from 1 to n p
n p2
E p 1.2.3....
n p2
n p2
....
n are not divisible by p. p
similarly, we get E p (n !)
n p
n p3
n ps
Where s is the largest natural number such that p s
n
ps
1
Illustration 46. Find the exponent of 3 in 100!. Solution:
In terms of prime factors 100 ! can be written as 2a. 3b 5c. 7d……………………. Now, E3 (100!)
100 3
100 32
100 33
100 34
=33 + 11+3+1=48 Hence, the exponent of 3 in 100 ! is 48 ( 100 ! = 2a. 348 .5c .7d )
E2 (33!)
33 2
33 22
33 23
33 24
33 25
= 16+8+4+2+1 = 31 Hence, the exponent of 2 in 33 ! in 31. Now 33 ! is divisible by 231 which is also divisible by 219 largest value of n is 31. Illustration 47. Find the number of zeros at the end of 100!? Solution:
In terms of prime factors 100 ! can be written as 2a. 3b .5c. 7d Now, E2 (100!)
100 2
100 22
100 23
100 24
100 25
100 26
50 25 12 6 3 1 97 100 5
and E5 (100!)
100 52
20 4 24 100! 297.3b.524.7 d ... 273.3b.(2 5) 24 .7 d ... 273.3b.(10) 24 .7 d ... Hence number of zeros at the end of 100! is 24. OR Exponent of 10 in 100! min (97, 24)
24.
Important Results to Remember (i)
n straight lines are drawn in the plane such that no two lines are parallel and no three lines are concurrent. Then the number of parts into which these lines divide the plane is equal to 1+
n
Illustration 48. Six straight lines are drawn in the plane such that no two lines are parallel and no three lines are concurrent. Then find the number of parts into which these lines divide the plane. Solution.
No. of parts of the plane = 1 + =1+
(ii)
6
6.7 = 22 1.2
The sum of the digits in the unit place of all numbers formed with the help of a1 ,a2………… an taken all at a time is = (n — 1)! (a1 ,a2………… an) (repetition of digits not allowed)
Illustration 49. Find the sum of the digits in the unit place of all numbers formed with the help of 3, 4, 5, 6 taken all at a time. Solution.
Sum of the digits in the unit place = (4—1)!(3+4+5+6) = 6.18 = 108.
(iii)
The sum of all digit numbers that can be formed using the digits a1 ,a2………… an (repetition of digits not allowed) is
(n 1)!(a1 a2 ............... an )
(10n 1) 9
Illustration 50: Find the sum of all five digit numbers that can he formed using the digits 1, 2,3, 4 and 5 (repetition of digits not allowed).
(105 1) 9
Solution.
Required sum= (5—1)! (1 +2+3+4+5) = 24.15. 11111 = 3999960.
Alternative Method: Since one of the numbers formed with the 5 digits a , b , c , d, e is 104a+ 103b + 102c+ l0d+e; Hence 104a will occur altogether in 4 ! Ways similarly each of 104b, 104c, 104d, 104e will occur in 4 ! Ways. Hence, if all the numbers formed with the digits be written one below the other, thus 104.a 103.b 102.c 10.d e 104.b 103.c 102.d 10.e a 104.c 103.d 102.e 10.a b 104.d 103.e 102.a 10.b c 104.e 103.a 102.b 10.c d Hence the required sum 4!
(a b c d e) (104 103 102 10 1)
4!
(1 2 3 4 5) (11111)
3999960.
(iv)
If there are n rows, I row has placed
1
squares. II row has
2
. squares, III row has
3
squares, and so on. If we
Xs in the squares such that each row contain at least one X. The number of ways = coefficient of
x in 1
C1 x
.......
1
3
C2 x 2 ............
1
C 1x
1
2
C1 x
2
C2 x 2 ............
2
C 2x
1
( 3 C1 x
3
C2 x 2 .....
C 3x 3)
Illustration 51. Six X’s have to be placed in the squares of the figure below, such that each row contains at least one X. In how many different ways can this be done?
Solution:
The required no. of ways = coefficient of x6 in ( 2 C1 x
2
C2 x 2 )( 4 C1 x
4
C2 x 2
= coefficient of x3 in (2 + x)2 (4 + 6x + 4x2 + x3)
4
C3 x 3
4
C4 x 4 )( 2 C1 x
2
C2 x 2 )
= coefficient of x3 in (4 + 4x + x2) (4 + 6x + 4x2 + x3) = 4+16+6 = 26
Alternative Method In the given figure there are 8 squares and we hay to place 6X ‘s this can be done in
X
8
i.e. C6
8
C2
8.7 1.2
X
X
X
X
X
28 ways
But these include the possibility that either headed row or lowest row may not have any X. These two possibilities are to be excluded. Required number of ways = 28 — 2 = 26
Solved Examples
Example 1: A letter lock consists of three rings each marked with 10 different letters. In how many ways is it possible to make an unsuccessful attempt to open the block? Solution:
Two rings may have same letter at a time but same ring cannot have two letters at time, therefore, we must proceed ring wise. Each of the three rings can have any one of the 10 different letters in 10 ways. Therefore Total number of attempts = 10 × 10 × 10 = 1000. But out of these 1000 attempts only one attempt is successful. Therefore Required number of unsuccessful attempts = 1000 - 1 = 999.
Example 2: Find the total number of signals that can be made by five flags of different colour when any number of them may be used in any signal. Solution: Case I :
When only one flag is used. No. of signals made = 5P1 = 5.
Case II : When only two flag is used. Number of signals made = 5P2 = 5.4 = 20. Case III : When only three flags are used. Number of signals is made = 5P3 = 5.4.3 = 60. Case IV : When only four flags are used. Number of signals made = 5P4 = 5.4.3.2 = 120. Case V :
When five flags are used. Number of signals made = 5P5 = 5! = 120. Hence, required number = 5 + 20 + 60 + 120 + 120 = 325.
Example 3: Prove that if each of the m points in one straight line be joined to each of the n points on the other straight line, the excluding the points on the given two lines. Number of points of intersection of these lines Solution:
is 1/4 mn (m-1(n-1).
To get one point of intersection we need two points on the first line and two points on the second line. These can be selected out of n-points in nC2 ways and for m points in mC2 ways. 2n 1 C0 Therefore Required number = Now
=
2n 1
C2 n
1
1
2n 1
C1
2n 1
C2 n etc.........
m m 1 n n 1 . 2! 2!
= 1/4 m n (m - 1)(n - 1) Example 4:
There are ten points in a plane. Of these ten points four points are in a straight line and with the exception of these four points, no other three points are in the same straight line. Find (i)
the number of straight lines formed.
(ii)
the number of triangles formed.
(iii) the number of quadrilaterals formed by joining these ten points. Solution: (i)
For straight line, we need 2 points
No. of point selected out
No. of points selected out
No. of straight line
of 4 collinear points
of remaining 6 points
formed
0
2
4
C0 × 6C2 = 15
1
1
4
C1 × 6C1 = 24
2
0
1
(In last case only one straight line is formed) Therefore Required number = 15 + 24 + 1 = 40 (ii)
For triangle, we need 3 points
No. of point selected out
No. of points selected out
of 4 collinear points
of remaining 6 points
No. of triangles formed
0
3
4
C0 × 6C3 = 20
1
2
4
2
1
4
3
0
C1 × 6C2 = 60 C2 × 6C1 = 36 0
(In last case number of triangles formed is 0) Therefore Required number = 20 + 60 + 36 + 0 = 116 (iii)
For a quadrilateral, we need 4 points
No. of point selected out
No. of points selected out
No. of quadrilateral
of 4 collinear points
of remaining 6 points
formed
0
4
4
C0 × 6C4 = 15
1
3
4
2
2
4
3
1
0
4
0
0
C1 × 6C3 = 80 C2 × 6C2 = 90
(In these cases number quadrilateral is formed Therefore Required number = 15 + 80 + 90 = 185. Example 5:
A student is allowed to select at most n-blocks from a collection of (2n + 1) books. If the total number of ways in which he can select a book is 63, find the value of n.
Solution:
Since the student is allowed to select at the most n-books out of (2n + 1) books, therefore, he can choose, one book, two books or at the most n books. The number of ways of selecting at least one books are 2n 1
C1
2n 1
.........2 n 1 Cn
C2
63
S (Say)
Again, we know that 2n 1
C0
2n 1
Now 2 n 1 C0
2n 1
C2 n
1 + 1 + 2S = 22n+1 or
.........2 n 1 Cn
C1
2 + 2 . 63 = 22n+1
1
1
2n 1
2n 1
C1
C2 n
2n 1
1
22 n
1
C2 n etc......... Hence, we have
or
128 = 22n+1 or 27 = 22n+1
=>
2n+1 = 7
or
2n = 6 n=3
Example 6:
A family consists of a grandfather, 6 sons and daughters and 5 grand children. They are to be seated in a row for dinner. The grand children wish to occupy the two seats at each end and the grandfather refuses to have a grandchild on either side of him. In how many ways can the seating arrangements be made for the dinner?
Solution:
There are 6 adults, 4 grand children and 1 grand father. Let us mark the seat for 11 persons from 1 to 11. S 1
S 2
S 3 .....................9
S
S
S
10
11
Seats number 1, 2 and 10, 11 at the ends are to be occupied by 4 grand children and it can be done in 4
P4 = 4! = 24 ways.
Now, we will seat the grandfather who cannot occupy seat number 3 or seat number 9 because he does not want to have a child by his side. Hence, he has to choose any of five seats 4, 5, 6, 7, 8 i.e. can seat himself in 5 ways. Example 7:
In an examination, the maximum marks for each of the three papers are 50 each. Maximum marks for the fourth paper is 100. Find the number of ways in which the candidate can score 60% marks in the aggregate.
Solution:
Aggregate of marks 50 × 3 + 100 = 250 Therefore 60% of the aggregate = 3/5 × 250 = 150 Now the number of ways of getting 150 marks in the aggregate = coefficient of x150 in (x0 + x1 +.........+ x50)3 (x0 + x1 + x2 +......+ x100) = coeff. of x150 in ((1 - x51)/(1 -x))3 ((1 - x101)/(1 - x)) = coeff. of x150 in (1-x51)3(1-x101)(1-x)-4 = coeff. of x150 in (1-3x51 + 3.x102 - x153)(1-x101)(1-x)-4 coeff. of x150 in 1 – 3.x51 – x101 3.x102 3.x152 ...
1 4 x x ( r 3)Cr x r
= 1.(151.152.153)/6 - 3(100.101.102)/6 - 1(50.51.52)/6 + 3(49.50.51)/6 = 151.76.51 - 100.101.51 - 50. 17.26 + 49.25.51 = 57 (151.76 - 100 - 101) + 17 (49.25 - 50.26) = 51 (11476 - 10100) + 17 (3675 - 1300) = 51.1376 + 17.2375 = 70176 + 40375 = 110551.
Example 8:
Find the number of 4 lettered words that can be formed form the letters of the word PROPORTION?
Solution: Step 1
Letter
Freq.
P
2
R
2
O
3
T
1
I
1
N
1
-------------------------------------------10 (always check total) Step 2 (i)
here we cannot have all the 4 letters alike, so
(ii) 3 alike Say X
1 diff.
Selection 1
Y
Remark
5
1
C1 → 3O's
C1 × C1 = 5
5
C1 → one form {P, R, T, I, N}
Arrangement of these 3X's and 1 Y =
4 4
P 3! 1!
= 4!/3! = 4 Hence total such words = 5 × 4 = 20. (iii) 2 same
2 diff.
2(X)
(Y, Z)
Selection 3
C1 × 5C2 = 12
Arrangement = (p+q+r)/p!q!r! = (2+1+1)2!1!1! = 12 Hence total words = 12 × 30 = 360 (iv) 2 same (2X)
2 another same (2Y)
Selection 3
C2 = 3
Arrangements = (2+2)!/2!2! = 6 Total words = 3 × 6 = 18 (v) 1 same, 3 diff º 4diff. Selection 6C4 = 6C2 = 15 Arrangement = (1+1+1+1)!/1!1!1!1! = 4! = 24 Total words = 15 × 24 = 360 (vi) 1 same, 3 another same º case ii (already considered) Hence grand total of desired words = 20 + 360 + 18 + 360 - 758.
Example 9:
You are given the responsibility of organizing a fresher's welcome party at IIT-Kanpur. Total 496 students will join IIT-K. Six restaurants A, B, C, D, E and F are booked. Capacity of each is 120, 146, 46, 72, 72, 80 persons at a time respectively. You have to group them in such a way that each group has least possible number of students. How would you do so?
Solution:
Remember: Equal division minimizes the number of students in each group. But: 496/6 = 82.6 > 46 Therefore C has least capacity of 46 and is not capable to accommodate 82 students. Hence let us group for it first i.e. 46 students get accommodated. Students left 450
Restaurant left A B D E and F
But 450/5 = 90 > 72 Next D and E each require smallest group 72, 72 D → 72 Students left 306
E → 72 Restaurant left A B and F
Now F is the least left for grouping, so assigning 80 to F. Students left 226
Restaurant left A and B
Now Therefore 226/2 = 113 < 120 and each of A and B is capable of accommodating to this. Hence, A → 113 B → 113 But not A → 120 B → 106 As in the case A does not have minimum possible number of students. Hence, we grouped 496 students into 46, 72, 72, 80, 113 and 113 each. No. of ways of doing so = 496! / (46!(72!)2 .2!80!(113!)2.2!) Example 10: How many different numbers can be formed with the digits 1, 3, 5, 7, 9 when taken all at a time and what is their sum? Solution:
The total number of numbers = |5 = 120. Suppose we have 9 in the unit's place. We will have |4 = 24 such numbers. The number of numbers in which we have 1, 3, 5 or 7 in the unit's place is |4 = 24. Hence, the sum of the digits in the unit's place in all the 120 numbers = 24 (1 + 3 + 5 + 7 + 9) = 600.
The number of numbers when we have any one of the given digits in ten's place is also |4 = 24 in each case. Hence, the sum of the digits in the ten's place = 24 (1 + 3 + 5 + 7 + 9) tens = 60 tens = 600 × 10. Proceeding similarly, the required sum = 600 units + 600 tens + 600 hundreds + 600 thousands + 60 ten thousands. = 60 (1 + 10 + 100 + 1000 + 10000) = 600 × 11111 = 6666600. Example 11: Find the values of r for which the number fo combinations of n things taken r at a time is greatest. Solution:
Since
n
Cr = (n(n-1)(n-2)....(n-r+2)(n-r+1))/(1.2.3...(r-1)r)
n
and
Cr-1 = (n(n-1)(n-2)....(n-r+2))/(1.2.3...(r-1))
Therefore
n
Cr = nCr-1 × (2-r+1)/r = nCr-1(((n+1)/r)-1).
The multiplying factor (n-r+1)/r may be written as (((n+1)/r) - 1), which shows that it decreases as r increases. Hence as r resumes the values 1, 2, 3,...... in succession, nCr is continually increased until (((n+1)/r) - 1) becomes equal to 1 or less than 1. Now (((n+1)/r) - 1) > 1 so long as (n+1)/r) - 2 that is (n+1)/r) - r . We have to choose the greatest value of r consistent with this inequality. (1) Let n be even, and equal to 2m, then (n+1)/2 = (2m+1)/2 = m+(1/2) ; and for all values r up to m (inclusive of m) this is greater than r. Hence by putting r = m = n/2 , we find that the greatest number of combination is nCn/2. (2) Let n be odd, and equal to 2m + 1; then (n+1)/2 = (2m+1)/2 = m+1 ; and for all values of r up to m (inclusive of m) this is greater than r, but when r = m + 1 the multiplying factor becomes equal to 1, and nCm+1 = nCm; that is n n+1
C/ 2 = n+1nC/ 2 and
therefore the number of combinations is greatest when the things are taken (n+1)/2 or (n-1)/2 at a time, the result being the same in the two cases.
