## Permutation & Combination Solved Questions For IBPS, SBI and Other Bank Exams - Page 1

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Permutation & Combination Solved Questions (Page 1 of 5)

Below you will find important formulas of permutation and combination problems. Following that you will find solved questions. These questions will help you in preparation for all bank exams including IBPS, SBI, RRB and other banks.

Permutations:
The different arrangements of a given number of things by taking some or all at a time are called permutations.

Important formulas:

1. Number of permutations of n different things, taken at a time r, where repetition is not allowed is denoted by nPr, and given by,
nPr = n! / (n – r)! Where 0≤ r ≤ n.

2. Number of permutations of n objects taken all at a time, where p1 things are of first type, p2 things are of second type,...,pk things are of kth type is given by,
n! / p1!p2!...pk!

3. Number of permutations of n distinct things taking them all at a time = nPn = n!
Note:
nP0 = 1

Combination:
Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

4. Number of combinations of n different things taken at a time r is denoted by nCr, is given by,
nCr = n! / r!(n – r)!, where 0 ≤ r ≤ n.

Note:
nC0 = 1 = nCn
nCr = nCn-r when 0 ≤ r ≤ n.
nCr-1 + nCr = n+1Cr
nCp = nCq if either p=q or p+q = n

Question 1

Find the number ways to arrange the letters in the word 'LETTER'.
a)20 b)360 c)180 d)5040

Solution :

The word 'LETTER' contains 6 letters, namely 1-L, 2-E, 2-T and 1-R.

Required number of ways = 6! / (1!)x(2!)x(2!)x(1!) = 720 / 4 = 180.

Question 2

In how many different ways can the letters of the word 'SYSTEMATIC' be arranged so that the vowels always come together?

a)10080 b)60480 c)362880 d)none of these

Solution:

In the word 'SYSTEMATIC', consider the vowels EAI as one letter.

Thus, we have SYSTMTC(EAI).

The word has 8 (7 + 1) letters of which S & T occurs 2 times and the rest are 1 time.

Number of ways of arranging these letters = 8! / (2!*2!) = 40320 / 4 = 10080.

Now, 3 vowels in which each are different, can be arranged in 3! = 6 ways.

Required number of ways = (10080*6) = 60480.

Question 3

How many 3-letter words can be formed with or without meaning from the letters M, Y, A, S, I and N , which are ending with M amd none of the letters should be repeated?
a)20 b)18 c)25 d)27

Solution :

Since each desired word is ending with M, the least place is occupied with M. So, there is only 1 way.

The second place can now be filled by any of the remaining 5 letters (Y, A, S, I, N). So, there are 5 ways of filling that place.

Then, the first place can now be filled by any of the remaining 4 letters. So, there are 4 ways to fill.

Required number of words = (1 x 5 x 4) = 20.

Permutation & Combination Solved Questions (Page 1 of 5)

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