Question

Three ladies have brought one child each for admission to a school. The principal wants to interview the six persons one by one subject to the condition that no mother is interviewed before her child. Then find the number of ways in which interviews can be arranged.

Answer 1

A child must interview first.

Then either her mother or another child must interview second.

$Case$ $1.$ Her mother interviews 2nd

Then another child must interview 3rd.

Then 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 child${}^{\prime }$s interview first and second.

$Sub-case$ $2a:$ A child interviews third.

The 3 mothers can interview in any of $3!$ or $6$ ways.

That's $6$ $ways$ for $sub-case$ $2a.$

$Sub-case$ $2b:$ A mother interviews third.

We can choose that mother in $2$ ways, as the first or second

interviewing child${}^{\prime }$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 there are$2\times 3=6$ $ways$ for $sub-case$ $2b.$

So there are $6+6=12ways$ for $case2.$

That's a total of $3+12=15ways$ for the scheme of interviewing.

Then there are $3!=6ways$ the three mother-child pairs can be

permuted.

$\therefore Totalways:15\times 6=90ways.$