Question

Find the number of ways in which \[5\] boys and \[3\] girls can be seated in a row so that each girl is in between two boys.
A. \[2880\]
B. \[1880\]
C. \[3800\]
D. \[2800\]

Answer 1

Hint: Here, in the given question, we have been asked to find the number of ways in which \[5\] boys and \[3\] girls can be seated in a row so that each girl is in between two boys. First we will find the number of ways that boys can be seated. After that, we will observe that there will be four places where girls can be seated. We will arrange them in the possible places to get the desired result.

Complete step-by-step answer:
Given: There are \[5\] boys and \[3\] girls
At first, we will arrange the \[5\] boys in a row which can be done in \[{}^5{P_5}\] ways,
Therefore, the boys can be arranged in \[5!\] = \[120\] ways.
Arrangement is done in the following way, where on each empty place, a girl can be seated.
 \[B \ldots B \ldots B \ldots B \ldots B\]
Clearly, there are \[4\] places where girls can be seated. And there are only \[3\] girls.
So, the number of ways that \[3\] girls can be arranged at \[4\] places are \[{}^4{P_3}\] ways = \[\dfrac{{4!}}{{1!}}\] ways
= \[4 \times 3 \times 2\] ways
= \[24\] ways.
It means, for each arrangement of boys, there are \[24\] ways of arranging girls.
Therefore, for \[120\] arrangement of boys, total number of ways of arranging girls = \[120 \times 24\] ways
= \[2880\] ways.
So, the correct answer is “Option A”.

Note: The number of permutations of \[n\] different things, taken \[r\] at a time, where repetition is not allowed, is denoted by \[{}^n{P_r}\] and is calculated by \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] , where \[0 \leqslant r \leqslant n\] .
Additional information: The notation \[n!\] represents the product of first \[n\] natural numbers. Therefore, the product \[1 \times 2 \times 3 \times \ldots \times \left( {n - 1} \right) \times n\] is denoted as \[n!\] . And \[0!\] is defined as \[1\] .