# Qantitative Aptitude – Permutation & Combination for IBPS PO/Clerk

## Qantitative Aptitude – Permutation & Combination for IBPS PO/Clerk

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Takshila Learning regularly offers you different blogs and articles on important topics, news, and information related to banking exams. In this article, ‘Permutations’ from Quantitative aptitude section has been explained.

## PERMUTATIONS SOLUTIONS

Important concept & formulas: –

Factorial Notation: Let n be any positive integer. Then, factorial n denoted by n! is defined as

n! = n(n-1)(n-2). . . . . . . .3.2.1

eg:- 6! = (6 * 5 * 4* 3 * 2 * 1)

= 720

&0! = 1

Permutations: The numbers of arrangements possible of a given word or anything by taking some or all at a time are called permutations possible of that word or thing.

Example: – All permutations (or arrangements) made with the letters x, y, z by taking two at a time are (xy, yx, xz, zx, yz, zy).

Numbers of permutations of n different things, taken r at a time is given by

nPr = n(n-1)(n-2). . .. . . (n-r+1)

= n! / (n-r)!

Note: If there are in total n objects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of the third kind and so on and pr are alike of rth kind, such that (p1+p2+. . . . . . . . pr) =n

Then, a number of permutations possible of these n objects are defined as:

n! / (p1!).(p2!). . . . .(pr!)

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Let us understand the concept of permutations with the help of some examples:

Example: Evaluate 30!/28!

Sol: 30!/28! = 30 * 29 * (28!) / (28!)

= 30 * 29 =870

Example: How many 4-letter words with or without meaning can be formed from the letters of the word ‘LOGARITHMS’ if repetition of the letters is not allowed.

Sol: As ‘LOGARITHMS’ contains 10 different letters

So, Number of permutations possible = Number of arrangements possible by 10different letters taking4 at a time

= 10P4

= 10 * 9 * 8 * 7

= 5040

Example: In how many ways can the letters of word ‘LEADER’ be arranged?

Sol: The word ‘LEADER’ contains 6 letters i.e., 1L, 2E ,1A ,1D and 1R

Required number of ways

= 6! / (2!)

= 6 * 5 * 4 * 3 * 2 *1 / 2 * 1

=360

Example: How many arrangements can be formed out of the letters of the word ‘MATHEMATICS’ so that the vowels always come together?

Sol: In the word ‘ MATHEMATICS’ we treat vowels AEAI as one letter thus we have MTHMTCS (AEAI)

Now we have to arrange 8 letters out of which M occurs twice, T occurs twice & the rest are different

Number of ways of arranging these letters

= 8! / (2!)(2!)

= 10080

Now AEAI has 4 letters in which A occurs 2 times and the rest are different

Number of ways of arranging these letters

= 4! / 2! = 12

Required number of words = (10080 * 12)

= 120960

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Learn PROBABILITY SOLUTIONS – Questions & Answers here..

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August 25, 2019