CONTENTS S.NO.
TOPIC
PAGE NO.
1.
INTRODUCTION
2
2.
FUNDAMENTALS OF COUNTING
2
3.
USE OF FACTORIAL
3
4.
PERMUTATIONS
5
5.
COMBINATIONS
6
6.
IMPORTANT METHODS USED IN PERMUTATIONS AND COMBINATIONS
12
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PERMUTATIONS AND COMBINATIONS
1. INTRODUCTION : In each of the test papers of JEE (Mains and Advanced) you’ll find 3-5 sums on based on Permutations and Combinations and Probability. 80% of Probability problems are based on Permutations and Combinations (P&C). So, unless the fundamentals of P&C are clear Probability cannot be handled.
Permutations and Combinations are not that straight forward, since the question is formed in a language which needs to be first understood and then be formulated into Mathematics. Once you get habituated to all the fundamentals and the problems, you can handle the problems with a good flow. This Chapter aims at clearing all the fundamentals required for solving all kinds of problems and give you an eyesight of observing the kind of problem and the method to solve it.
2. FUNDAMENTALS OF COUNTING: If an operation A can be performed in ‘n’ different ways and if corresponding to each of these ways there are ‘m’ different ways of performing another operation B, then both the operations can be performed in n x m different ways. Ex. 1.1.
Entrance and Exit: How many different ways are there for entering from entrances and exiting from 3 exits? If there are 4 entrances and 3 exits then there can be 4 x 3 = 12 different ways for entering and exiting.
Soln:
Ex. 1.2.
Path Travel: Let A, B and C be three cities and from A to B there are m different routes and from B to C there are n different routes. Then the number of different paths one can traverse in going from A to C are?
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Soln:
Suppose a person travels from A to B by route 1 then at B he has n choices of paths to reach C. Therefore, for every path from A to B he has n paths at his disposal. So, total number of routes he can undertake is n x m.
Ex. 1.3.
Even Natural Numbers: There are how many 3 digit even natural numbers?
Soln:
Since the number should have 3 digits, the first digit i.e. the hundreds digit should not be 0. Therefore hundreds place can be filled in 9 ways (1 to 9). The tens digit has no restriction and can be filled in 10 ways (0-9). Since the number should be even number we can only fill the last position with these numbers: 0, 2, 4, 6 and 8. Therefore, the total number of 3 digit even natural numbers are 9x10x5= 450.
3. USE OF FACTORIAL: I.
Basic Principle: The product of all consecutive integers starting from 1 to n is denoted by n! and read as ‘n factorial’. n! = 1 x 2 x 3 x 4 x … x (n-2) x (n-1) x n 0! = 1
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Following values should be memorized for fast calculations: Number 1 2 3 4 5 6 7 8 9 10
Factorial Value 1 2 6 24 120 720 5040 40320 362880 3628800
II.
Linear Arrangements:
a.
All elements different: Suppose we are given a number of different objects (Let’s say a, b and c) and asked in how many different ways can we arrange them in a straight line? According counting principle the number of arrangements possible are = 3 x 2 x 1 = 6 (3!), for four different objects 4 x 3 x 2 x 1 = 24 (4!) and so on. This fundamental will appear clearer by going through following examples:
Ex. 2.2.1.
How many different words can be formed from the letters of the word ‘Bread’?
Soln:
5 x 4 x 3 x 2 x 1 =120 = 5! b.
Some elements identical: i. Suppose we are given a set of elements: a, a, b and asked - In how many different was can we arrange them in a straight line? 3 arrangements are possible: aab, baa, aba, which also equal to 3!/2!. ii. For a, b, c and c, there are 12 possible arrangements: abcc, acbc, accb, cacb, ccab, bacc, bcac, bcca, ccba, cbca, cabc and cbac, which is also equal to 4!/2!. iii. For a,b,b,c,c,c number of possible arrangements= 6!/(2!.3!) We can obtain a generalized rule for some same objects. Let’s say there are total of n objects, in which there are p number of objects of one kind, q of one kind and r of one kind. Then the number of arrangements possible is:
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This can be extended to any number of same kind of objects. Ex. 2.2.2.
How many different words can be made from the letters of the word: MISSISSIPPI? Here we have four Is, four Ss and 2 Ps. So, denominator will have 4! (for 4 Is), 4! (for 4 Ss) and 2! (for 2 Ps). Therefore, the answer is: 11! / (4! 4! 2!)= 34650.
Soln:
4. PERMUTATIONS: All the arrangements that can be made by taking some or all the items is called a Permutation. Permutation implies “arrangement” where order is of importance. The number of permutations of n things taking r at a time is denoted by nPr,
Some useful valurs:
Fundamental of permutation will appear clearer by going through following examples: Ex.3.1.
How many ways 5 men can be seated in 3 chairs?
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Soln:
Here, out of 5 men we have to arrange 3 men the in 3 chairs. Therefore, number of permutations of 5 people taken 3 at a time are:
Ex. 3.2. Soln:
How many ways 5 men can be seated in 1 chair?
Ex. 3.3.
How many four letter words can be formed using the letters of the word “Roaming”? None of the letters in the word are repeated. Therefore, the number of four letter words that can be formed:
Soln:
7P4 = 840
5. COMBINATIONS:
a.
Explanation:
Each of the groups or selections which can be made by taking some or all of a number of items is called a Combination. In combinations, the order in which the items are taken is not considered as long as the specific things are included. Following example will give a clear idea about combinations:
Ex. 4.1. Soln:
How many combinations of three items a, b and c are possible, taken two at a time? Here ab and ba are not considered separately because the order in which a and b are taken is not important but is only required that a combination including a and b be counted. Therefore, possible combinations are: : ab, bc and ca i.e. 3 combinations.
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The number of combinations of n things taking r at a time is denoted by nCr (and read as “nCr”). Let the number of combinations nCr be m. Consider on these m combinations. Since this is a combination, the order of the r itmes is not important. If we now impose the condition that order is required for these r items, we can get r! arrangements from this one combination. So each combination can give rise to m x r permutations. But since these are al permutations of n things taken r at a time, this must equal to nPr. So,
Hence,
This formula gives the number of ways in which r things can be selected from n things. b.
Important Results:
This means just that the number of ways of choosing r objects out of n objects is the same as the number of ways of not choosing (n – r) objects out of n objects. In the Paths travelled example the number of ways of selecting r paths out of n paths is same as rejecting (n-r) paths out of n.
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5. If n is even, then the greatest value of nCr is nCm, where m = n/2. If n is odd, then the greatest value of nCr is nCm, where m = (n-1)/2 or (n+1)/2. c.
Binomial Theorem:
Binomial theorem describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the power (x + y)n where
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n is positive integer and the coefficient of each term is a specific positive integer depending on n. For example,
Generalizing the expansion of (x + y)n into a sum:
Where
is nCk.
Simple and useful variation of this expansion:
Or equivalently,
Many forms can be generated by entering the values of x and n in the above expression. Some of the useful substitutions are: x=1 and x = -1 where n can have any value. For x=1, n=4, (1 + x)4 = 24 = 4C0.14 + 4C1.13 + 4C2.12 + 4C3.11 + 4C4.10
For x = -1, n = 4, (1 + x)4 = 0 = 4C0.14 - 4C1.13 + 4C2.12 - 4C3.11 + 4C4.10
Therefore, 4
C0.14 + 4C2.12 + 4C4.10 = 4C1.13 + 4C3.11. You may come across such variations
in exam.
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Following values can be useful in solving problems:
1. If n = 2m,
d. Pascal’s Triangle:
Given by a French Mathematician named Blaise Pascal
Significance of Pascal’s Triangle:
Row Number (n)
Values
0
0
C0 = 1
1
1
C0 = 1, 1C1 = 1
2
2
C0 = 1, 2C1 = 2, 2C2 = 1
3
3
C0 = 1, 3C1 = 3, 3C2 = 3, 3C3 = 1
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4
C0 = 1, 4C1 = 4, 4C2 = 6, 4C3 = 4, 4C4 =1 5 C0 = 1, 5C1 = 5, 5C2 = 10, 5C3 = 10, 5 C4 = 5, 5C5 = 1
4 5
The first row consists of single element 0C0 = 1. The second row consists of 1C0 and 1C1. From the third row onwards, the rule for writing the entries is as follows: Add consecutive elements in the previous row and write the answer between the two entries. After exhausting all possible pairs, put the number 1 at the beginning and the end of the row. For example, the third row is got as follows. Second row contains only two elements and they add up to 2. Now, put 1 before and after 2. For the fourth row, we have 1 + 2 = 3, 2 + 1 = 3. Then, we put 1 + 2 = 3, 2 + 1 = 3. Then we put 1 at the beginning and the end.
Usage of Pascal’s Triangle:
The binomial coefficients 1, 2, 1 appearing in this expansion correspond to the third row of Pascal's triangle. The coefficients of higher powers of x + y correspond to later rows of the triangle:
The powers of x go down until it reaches 0, starting value is n. The powers of y go up from 0 until it reaches n. The nth row of the Pascal's Triangle will be the coefficients of the expanded binomial. For each line, the number of products (i.e. the sum of the coefficients) is equal to 2n. For each line, the number of product groups is equal to n+1. Evaluate the expansion of (2x – y)4
Ex.4.2. Soln:
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6.
IMPORTANT METHODS OF ARRANGEMENT
a. Circular Arrangement:
Circular arrangement will lead to the same arrangements for a circular table. Hence one circular arrangement corresponds to n unique row (linear) arrangements. Hence the total number of circular arrangements of n persons is = (n - 1)!. In other words the permutation in a row has a beginning and an end, but there is nothing like beginning or end in circular permutation. Hus, in circular permutation, we consider one object is fixed and the remaining objects are arranged in (n - 1)! ways (as in the case of arrangement in a row). Distinction between clockwise and Anti-clockwise Arrangements Consider the following circular arrangements:
In figure I the order is clockwise whereas in figure II, the other is anti-clock wise. These are two different arrangements. When distinction is made between the clockwise and the anti-clockwise arrangements of n different objects around a circle, then the number of arrangements = (n - 1)!. But if no distinction is made between the clockwise and anti-clockwise arrangements of n different objects around a circle, then the number of arrangements is (n - 1)!. b. String method
Suppose we have to arrange 6 persons in a row. They can be arranged in 6! ways. If two particular persons are tied together and consider it one. Then the numbers of persons = 6-2+1 = 5 which can be arranged in 5! ways and 2 persons which are tied can be arranged in 2! ways. So required number of arrangements = 5 ! x 2 ! = 240. c. Gap Method:
Suppose 4 men A, B, C, D are to be arranged in a row as xAxBxCxDx there will be 5 gaps between these 4 men. 3 in between and 2 at either end. If three women are to be arranged so that they are never together we shall use gap
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method i.e. we have to arrange them in between these 5 gaps. Hence no. of arrangements are 5P3. d. String method:
Suppose we have to arrange 6 persons in a row. They can be arranged in 6! ways. If two particular persons are to be always together then we can consider them as if they are tied together and consider them as one entity. Then the numbers of persons = 6-2+1 = 5 which can be arranged in 5! ways and 2 persons which are tied can be arranged in 2! ways. So required number of arrangements = 5 ! x 2 ! = 240. e. When two persons are never together:
= total - together = 720 - 240 480. f.
Selection of atleast one man for the committee out of men and women:
Total - no men = Selection of committees with atleast one. Similarly, at least two men = Total - (no man + one man) g.
Restricted Selection and Arrangement:
a. The number of ways in which r objects can be selected form n different objects if k particular objects are (i) Always included = n-kCr-k. (ii) Never included = n-kCr. b. The number of arrangement of n distinct objects taken r at a time so that k particular objects are: (i) Always included = n-kCr-k.r!, (ii) Never included = n-kCr.r!. Following example will give a clearer idea about this method: A team of four students is to be selected form a total of 12 students. In how many ways can the team be selected if,
Ex. 4.3.
1. 2. 3. 4.
All the students wish to be in the team. Two particular students have to be included in the team. Two particular students do not wish to be together in the team. Two particular students wish to be included together only,
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(a) Formation of team means selection of 4 out of 12. Hence the number of ways = 12C4 = 495.
Soln.
(b) Two particular students are already selected. Hence we need to select 2 out of the remaining 10. Hence the number of ways = 10C2 = 45. (c) The number of ways in which both are selected = 45. Hence the number of ways in which the two are not included together = 495 - 45 = 450. (d) There are two possible cases (i)
Either both are selected. In this case the number of ways in which this selection can be made = 45.
Or (ii)
both are selected. In this case all the four students are selected from the rest of ten students.
This can be done in 10C4 = 210 ways. Hence, the total number of ways of selection = 45 + 210 = 255.
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