Permutation and Combination with Explanation

Q8. In how many ways the 12 articles may be distributed among 4 persons so that each may get them in equal number.

Answer- Option A

Explanation-

12 articles when distributed among 4 persons, each will get 3 articles with each person can be arranged in 3! ways. So, total number of ways will be:

12! 3! 3! 3! 3!

=

12x11x10x9x8x7x6x5x4x3x2x1 3x2x1 x 3x2x1 x 3x2x1 x 3x2x1

= 3,69,600.

Q9. Evaluate ⁶P₂

Answer- Option B

Explanation-

⁶P₂ =

6! (6 - 2)!

=

6! 4!

=

6 x 5 x 4 x 3 x 2 x 1 4 x 3 x 2 x 1

= 30.

Q10. How many words can be formed by using all the letters of the word 'DAUGHTER' so that the vowels always come together?

Answer- Option A

Explanation-

Given word contains 8 different letters. When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter. Then, the letters to be arranged are DGHTR(AUE).

These 6 letters can be arranged in ⁶P₆ = 6! = 720 ways.
The vowels in the group (AUE) may be arranged in 3! = 6 ways.
Required number of words = (720 x 6) = 4320.

Q11. In how many ways can 21 boys and 19 girls stand in a queue such that now to girls can stand together?

Answer- Option C

Explanation-

In order that two girls are never together, we must place all these students as under:
G B G B G B G.... G B G
where B denotes the position of boys and G that of girl.
Since there are 21 boys, the number of places marked G are therefore, 22.

Now, 19 places out of 22 can be chosen in ²²C₁₉ = ²²C₃

22 x 21 x 20 3 x 2 x 1

= 1540.

Hence, the required number of ways = 1540.

Q12. How many different ways the letters of the word 'MACHINE' be arranged so that the vowels may occupy only odd positions?

Answer- Option B

Explanation-

There are 7 letters in the given word, out of which there are 3 vowels and 4 consonants.
Let us mark the positions to be filled up as follows:
(1) (2) (3) (4) (5) (6) (7)
Now, 3 vowels can be placed at any of the three places, out of the four marked 1, 3, 5, 7.
Number of ways of arranging the vowels = ⁴P₃ = (4 x 3 x 2) = 24.
Also, the 4 consonants at the remaining 4 positions may be arranged in ⁴P₄ = 4! = 24 ways.
Required number of ways = (24 x 24) = 576.

Q13. In how many ways can word with or without meaning, can be formed by using all the letters of the word 'DELHI,' using each letter exactly once?

Q14. How many words can be formed by using all letters of the word 'BIHAR'?

Answer- Option C

Explanation-

The word BIHAR contains 5 different letters.
Required number of words = ⁵P₅ = 5 ! = (5 x 4 x 3 x 2 x 1) = 120.