Question

Three ladies have each brought a child for admission to a school. The head of the school wishes to interview six people one by one, taking care that no child is interviewed before his/her mother. In how many different ways can the interviews be arranged?
A. $6$
B. $36$
C. $72$
D. $90$

Answer 1

This problem can be solved in two cases. We will first calculate the number of ways the interview can be arranged when the mother is interviewed second and a child is interviewed first. This will be our case one. In case two we will calculate the number of ways the interview can be arranged when two children are interviewed first and second. Now we will calculate the sum of two cases and declare that as the number of ways to arrange the interview.

Complete step-by-step answer:
We are given that three ladies have each brought a child for admission to a school.
Let us consider the different cases for the possible arrangement of the interviews as below,
Case - 1:-
The child is interviewed first.
The child’s mother is interviewed second.
Then another child is interviewed third.
Then we get that the last $3$ interviewees can occur $3$ ways.
$MCM$ which can occur only one way, or $CMM$ which can occur $2$ ways.
That’s $3$ ways for case $1$.
Case - 2:-
Two children are interviewed first and second.
Sub – case $2a$: A child is interviewed third.
We get that the $3$ mothers can be interviewed in any of $3!$ or $6$ ways.
That’s $6$ ways for sub – case $2a$.
Sub – case $2b$: A mother is interviewed third.
We can choose that mother in $2$ ways, as the first or second interviewing child’s mother.
Then for each of those $2$ ways, the last $3$ interviewees can occur $3$ ways. That is,
$MCM$ which can occur only one way, and $CMM$ which can occur $2$ ways.
So, we got that there are $2\times 3=6$ ways for sub – case $2b$.
So, we have $6+6=12$ ways for case $2$.
Adding them up, we have a total of $3+12=15$ ways for the scheme of interviewing.
Then there are $3!=6$ ways the three mother-child pairs can be permuted.
$\therefore$ We get the total ways as $15\times 6=90$ ways.

So, the correct answer is “Option D”.

Note: We can also find the solution for this in another method, we can arrange $6!=720$ ways the interview six people with no constraint. Divide by $2$ to account for ordering for each child/mother.
(e.g. given an ordering of six people, there is exactly one “equivalent” ordering by switching mother w/s his child).
Hence the answer is
$\begin{aligned} & =\dfrac{6!}{{{2}^{3}}} \\\ & =\dfrac{720}{8} \\\ & =90 \\\ \end{aligned}$
From both the methods we get the same answer.