Permutation and Combination Aptitude Questions and Answers

Explanation: The word 'NOMINATION' contains 10 letters, namely
3N, 2O, 1M, 2I,1A, and 1T
Required number of ways = 10 ! / (3!)(2!)(1!)(2!)(1!)(1!)
= 151200

Explanation: There are 7 different letters in the word 'ABOLISH'.
Therefore, The number of arrangements of any 4 out of seven letters of the word = Number of all permutations
of 7 letters, taken 4 at a time =
nPr = n(n - 1)(n - 2) ... (n - r + 1)
Here, n = 7 and r = 4, then we have
7p4 = 7 x 6 x 5 x 4 = 840.
Hence, the required number of ways is 840.

Explanation: The word 'POTENCY' has 7 different letters.
When the vowels EO are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTNCY (EO).
Now, 6 (5 + 1 = 6) letters can be arranged in 6! = 720 ways.
The vowels (EO) can be arranged among themselves in 2! = 2 ways.
Required number of ways = (720 x2)
= 1440.

Explanation: The word 'SPORADIC' contains 8 different letters.
When the vowels OAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters SPRDC (OAI).
Now, 6 letters can be arranged in 6! = 720ways.
The vowels (OAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (720 x 6) = 4320.

Explanation: The arrangement is made in such a way that the vowels always come together.
i.e., ""ZYMGN(OE)"".
Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ways.
The vowels ""AE"" can be arranged themselves in 2! ways; i.e.,2! = 2 ways
Therefore, required number of ways = 720 x 2 = 1440 ways

Explanation: There are 3 consonants and 3 vowels in the word DILUTE.
Out of 6 places, 3 places odd and 3 places are even.
3 vowels can arranged in 3 even places in 3p3 ways = 3! = 6 ways.
And then 3 consonants can be arranged in the remaining 3 places in 3p3 ways = 3! = 6 ways.
Hence, the required number of ways = 6 x 6 = 36

Explanation: The first bangle can be worn in any of the 2 hands (2 ways).
Similarly each of the remaining 11 bangles also can be worn in 2 ways.
Hence total number of ways=2×2×2×2×2×2×2×2×2×2×2×2
=2^12
=4096

Explanation: 6 subjects can be arranged in periods in 7P6 ways.
Remaining 1 period can be arranged in 6P1 ways.
Two subjects are alike in each of the arrangement.
So we need to divide by 2! to avoid over counting.
Total number of arrangements = (7P6 x 6P1)/2!
= 5040 × 6 / 2
= 30240 / 2
= 15120

Explanation: Required number of ways = 12C10 x 10C5
= 66 × 252
= 16632

Explanation: Therefore , required number of ways = 8C7 = 8C1
= 8

Explanation: 1 violet ball can be selected is 8C1 ways.
1 purple ball can be selected in 6C1 ways.
1 magenta ball can be selected in 4C1 ways.
Total number of ways = 8C1 × 6C1 × 4C1
= 8×6×4
= 192

Explanation: He has 15 patterns of cup and 9 model of saucer
A cup can be selected in 15 ways.
A saucer can be selected in 9 ways.
Hence one cup and one saucer can be selected in 15×9 ways =135 ways

Explanation: Number of arrangements possible = (6-1)!
= 5!
= 5×4×3×2×1
= 120

Explanation: Assume that in a party every person shakes hand with every other person
Number of hand shake = 66
Total number of hand shake is given by nC2
Let n = the total number of persons present in the party
nC2 =66
n (n-1) / 2 = 66
n^2 - n = 2 × 66
n^2 - n – 132= 0
n= 12 , -11
But we cannot take negative value of n
So, n = 12
Therefore number of persons in the party = 12

Explanation: Number of arrangements possible = {1} / {2} X (10-1)!
= {1} / {2}X 9!
= {1} / {2} X 9×8×7× 6×5×4×3×2×1
= {1 / 2 } ×362880
= 181440

Explanation: Three particular persons should not be included in each team.
i.e., we have to select remaining 6- 3= 3 persons from 12-3 = 9 persons.
Hence, required number of ways = 9C3
= {9×8 × 7} / {3 × 2 × 1}
= 504 / 6
= 84

Explanation: Here n = 7, k = 8. Hence, required number of ways = n^k =7^8

Explanation: Since each desired word is ending with G, the least place is occupied with G.
So, there is only 1 way.
The second place can now be filled by any of the remaining 5 letters (A , M , D , N , J ). 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.

Explanation: The word ‘WAGGISH’ contains 7 letters of which 1 letter occurs twice = 7! / 2!
= 2520
No. of permutations possible with vowels always together = 6! * 2! / 2!
= 1440 / 2
= 720
No. of permutations possible with vowels never together = 2520-720
= 1800.

Explanation: The word 'CABIN' has 5 letters and all these 5 letters are different.
Total number of words that can be formed by using all these 5 letters
= 5P5
= 5!
= 5×4×3×2×1
= 120