Permutation and Combination with Explanation

Q1. How many words can be formed with the letters of the word MUKESH such that every word begins with a vowel and ends with a vowel?

Answer- Option B

Explanation-

There are in all 6 letters in the word MUKESH, out of which 2 are vowels.
If each word is begin with a vowel and also end with vowel, there are remaining four letters which can be arranged in between, and also the two vowels can be interchanged.
Number of words formed as required
= 4! x 2! = 4 x 3 x 2 x 1 x 2 x 1 = 48.

Q2. Six persons enter a railway compartment in which there are nine seats. In how many ways they take their places? If there are only seven seats for them, is there any change in the number of arrangements?

Answer- Option A

Explanation-

There are 9 seats in all, and the passengers are 6 only.
Number of ways of taking seats will be
⁹P₆ = 9 x 8 x 7 x 6 x 5 x 4 = 60480.
If the number of seats are 7 only, the number of ways will be
⁷P₆ = 7 x 6 x 5 x 4 x 3 x 2 = 5040.

Q3. Calculate permutations of the letters of the following word SOFTWARE?

Q4. In how many ways the word ARTICLE may be arranged, when the vowels are evenly placed in the word.

Q5. In how many ways the word MANAGEMENT can be made out of the letters of the word?

Answer- Option B

Explanation-

There are total 10 letters in the word MANAGEMENT, out of which there are (2 M), (2 N), (2 A), (2 E) and G, T occur only once. Therefore, required number will be:

10! 2! 2! 2! 2!

=

10x9x8x7x6x5x4x3x2x1 2x1 x 2x1 x 2x1

= 226800.

Q6. In how many ways three Americans, three Indians, and three Japanese may sit around a round table in such a way that persons of the same nationality sit together?

Q7. In how many ways a cricket eleven may be selected out of 15 players, if (i) one specific player is to be selected in the team; (ii) one specific player is not to be included in the team.

Answer- Option D

Explanation-

(i) Out of 15 players, 11 are to be selected but one particular player is a must, so out of the remaining 14 only 10 players are to be selected. This can be done in ¹⁴C₁₀ ways.
¹⁴C₁₀ =

14! (14 - 10)! 10!

=

14! 4! 10!

=

14 x 13 x 12 x 11 x 10! 4 x 3 x 2 x 1 x 10!

= 1001.

(ii) One specific player is always to be excluded, so out of the remaining 14 players, 11 are to be selected. This can be done in ¹⁴C₁₁ ways.
¹⁴C₁₁ =

14! (14 - 11)! 11!

=

14! 3! 11!

=

14 x 13 x 12 x 11! 3 x 2 x 1 x 11!

= 364.