Question

The number of ways in which 4 boys and 4 girls can stand in a circle so that each boy and each girl stand one after the other, is:
$
  {\text{A}}{\text{. 3!4!}} \\
  {\text{B}}{\text{. 4!4!}} \\
  {\text{C}}{\text{. 8!}} \\
  {\text{D}}{\text{. 7!}} \\
$

Answer 1

Hint: To find the number of possibilities we first arrange the girls. Then we arrange the boys, this is ordered sampling with replacement (also known as permutation).

Complete Step-by-Step solution:
Given data, 4 boys and 4 girls alternately
We know n objects can be arranged in n! ways, (n! = n (n-1) (n-2)…….(n-n))

Let’s arrange the girls first.
If one girl takes a seat, then 3 girls and 3 seats are left.
There will be 3! ways in which the other 3 girls can take the seats.

Once the girls are seated, 4 boys and 4 seats are left.
 There will be 4! ways in which the 4 boys can take their seats.

So, the total number of ways = 3!4!
Even if we arrange a boy first, the total number of ways becomes = 4!3!, which is the same.
Hence Option A is the correct answer.

Note: The key in solving such types of problems is identifying that this question is an ordered sampling with replacement (permutation). In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. We can either arrange the boys or girls first, the process remains the same and gives the same result.