# Permutations and Combinations

**Topics:**Line, Numerical digit, Negative and non-negative numbers

**Pages:**12 (4398 words)

**Published:**April 15, 2013

Permutations and combinations is a very important chapter for many entrance examinations. Understanding of his topic requires student’s aptitude and through knowledge. In this chapter we will study different techniques of counting. Permutation and combination is nothing but “counting without counting” or counting by using different models. For example if we have to count total number of stars in the following grid: We do not need to count all the stars, as there are 3 rows and 4 columns, hence total number of star points are 4x3 = 12. Though this is a very simple example to explain fundamental principles of counting, it gives the basic introduction to counting. Now consider the second example, we have to find all three letter words which can be formed by using A, B, C with out repetition, The diagram indicates that the first place can be filled in 3 different ways, when first place is filled second can be filled in 2 different ways and then third place in 1 way. Hence total number of words is 3x2x1 =6 , the six words so formed are: ABC, ACB, BAC, BCA, CAB and CBA 1.1 FUNDAMENTAL PRINCIPLE OF COUNTING: As discussed in the above example, we can conclude that if event A can be done in ‘m’ ways and a second event B can be done independently in ‘n’ ways, then both the events A and B can be done in mxn ways. If events A and B are mutually exclusive then both the events can be done in m+n ways. Exclusive events are thoses which can not happen simultaneously. For example if a thief can enter the room through 4 doors or 5 windows, then total number of ways in which he can enter the room is not 4 x 5 but 4+5 as both cases are mutually exclusive. FACTORIAL NOTATION: Factorial is represented by the symbol ‘!’, n! =1.2.3.4….(n-1). It is always useful to have a simple notation for big products like 1x2x3x4x…..x99x100 = 100!. For example 4! =4.3.2.1 =24, 5! =5.4.3.2.1 =120, 0! =1 PERMUTATIONS: permutations are different no of arrangements. Making words of 4 letters without repetition or with repetition is an example of permutation. Now suppose we have to make a word of 4 letters by taking any 4 letters from 10 letters A, B, C, D, E, F, G, H, I __ __ __ __

10 9 8 7

Four places can be filled in 10x9x8x7 ways, which can be written as , which means and that is written as 10P4. So nPr = = n.(n-1).(n-2)….(n-r+1)

For example permutations of 4 distinct objects out of 20 distinct objects can be written as 20P4 Example: How many integers between 100 and 999 consist of distinct odd digits? Solution: We have to form all three digit numbers having distinct and odd digits. There are 5 odd digits 1,3,5,7 and 9. Total number of ways = 5x4x3 =60 ways. COMBINATIONS: A combination is a grouping or selection of all or part not a number of things without reference to arrangement of the things selected. Selection of three members from a family of ten, number of cricket matches in a group having 10 teams, number of triangles formed by joining any three points out of 20 points in plane etc. all are example of combination. The symbol nCr represents total number of selections taking r different objects out of n different objects. nCr =

Suppose there are 5 students and a company wants to select any 3 of them. So the total number of ways = selection of 3 out of 5 = rejection of 2 out of 5. Number of ways = 5C3 =5C2. or in general,

nCr = nCn-r

: Suppose there are n+1 students and we have to select r students out of these students. Now there are two cases: Case 1: a particular student X is always selected, total number of ways of selecting remaining r-1 students from the remaining n students = nCr-1 Case 2: a particular student X is never selected, total number of ways of selecting r students from the remaining n students = nCr Hence total number of ways = nCr-1 + nCr

n+1Cr = nCr + nCr-1 Example 2. In how many ways we can make two group of 10 and 3 students out of a total 13...

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