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 r^th 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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