1000+ Permutation and Combination Questions for CAT - 1

Ans: D

Number of letters in the word ‘KURUKSHETRA’ is 11 of which 2 are K’s. 2 are U’s, 2 are R’s and remaining are different.

∴ Required number of permutations = $${11!} / {2! 2! 2!}$$

= 4989600.

Ans: B

Required number of ways = $${7!} /{3!}$$ = 840.

Ans: C

There are 6 letters in the word ‘ATTEND’ whereas, T comes 2 times.

So, required number of ways = $${6!} / {2!}$$ =$${ 720} / {2}$$ = 360.

Ans: B

Total number of letters is 7, and these letters can be arranged in 7! Ways = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040 ways.

There are 7 letters in the world THERAPY including 2 vowels. (E, A) and 5 constants.

Consider 2 vowels as 1 letter

We have 6 letters which can be arranged in ⁶P₆ = 6 ways.

But vowels can be arranged in 2! Ways.

Hence, the number of ways, all vowels will come together

= 6! × 2!

= 1 × 2 × 3 × 4 × 5 × 6 × 2 = 1440

Total number of ways in which vowels will never come together

= 5040 - 1140 = 3600.

Ans: B

Required number of ways = $${7!}/{2!}$$ = 2520