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

Hint: Use the method of permutation and combination. A permutation is the number of ways in which objects from a set may be selected, generally without replacement to form a subset, whereas combination is the method of selection where the order of selection is a factor.

Complete step-by-step answer:
In this question, we need to determine the number of ways in which the interviews can take place such that first the interview of the child take place and then only for the mother. For this we need to follow the concept of combinations by considering C for children and L for their mothers.
Let C1, C2, C3 be the number of child and L1, L2, L3 being their mothers
Given that a child must be an interview before their mother
Now let us assume a pair of a child and a mother C1, and L1 is interview first, where child C1 is interview first, given as
\[\underline {{C_1}} \underline {} \underline {} \underline {} \underline {} \underline {} \]
Here lady L1 will be interviewed after the children, so the total number of possible ways lady L1 will be interviewed
\[
  {L_1} = {}^5{C_1} + {}^4{C_1} + {}^3{C_1} + {}^2{C_1} + {}^1{C_1} \\
   = 5 + 4 + 3 + 2 + 1 \\
   = 15ways \\
   = {}^6{C_2} \\
 \]
Now there are 4 slots left where C2, C3, L2, L3 will be interviewed; let’s consider the pair C2, L2 where C2 will be interviewed first; hence we have total 3 slots left for L2 to be interviewed
\[\underline {{C_2}} \underline {} \underline {} \underline {} \]
Hence the total number of possible ways to interview lady L2 will be
\[
  {L_2} = {}^3{C_1} + {}^2{C_1} + {}^1{C_1} \\
   = 3 + 2 + 1 \\
   = 6ways \\
   = {}^4{C_2} \\
 \]
Now there are only two slots left to interview C3 and L3, where the child will be interviewed first; hence the lady has only 1 slot left to be interviewed,
\[\underline {{C_3}} \underline {{L_3}} \]
Hence the total number of possible ways to interview lady L3 will be
\[
  {L_3} = {}^1{C_1} \\
   = \dfrac{{1!}}{{0!}} \\
   = 1way \\
   = {}^1{C_1} \\
 \]
Hence the total number of ways that the interview can be arranged
  \[
  Total = {}^6{C_2} \times {}^4{C_2} \times {}^1{C_1} \\
   = 15 \times 6 \times 1 \\
   = 90ways \\
 \]

Note: In this question, students must take care that whenever an interview will be conducted, the child first then only the ladies, vice-versa does not hold true here in this case.