Question

Four couples (husband and wife) decide to form a committee of four members. Find the number of different committees that can be formed in which no couple finds a place.

Answer 1

Hint: The different ways in which objects from a set can be picked, often without replacement, to produce subsets are done by applying different combinations to them. Combinations are applied for a set when one subset is formed with respect to choosing or not choosing an element from the set.

Formula Used: \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], where, C is a notation of combination, n is the total number of objects and r is the number of objects to be selected.

Complete step-by-step solution:
We have the given that out of four couples, that is 8 members, 4 members are to be selected in such a way that no two members are a couple in the committee.
We will create different cases for selecting the members of a committee by selecting 4 husbands, 3 husbands, 2 husbands, 1 husband, and 0 husbands and not selecting the wife with respect to the husband who is selected.
Case 1: First, we will select only the husbands out of 4 couples for the committee. The number of way to select 4 husbands out of 4 husbands and no wife is,
\[
  ^4{C_4}{ \times ^4}{C_0} = \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}} \times \dfrac{{4!}}{{0!\left( {4 - 0} \right)!}} \\
  { \Rightarrow ^4}{C_4}{ \times ^4}{C_0} = \dfrac{{4!}}{{4!0!}} \times \dfrac{{4!}}{{0!4!}} \\
  { \Rightarrow ^4}{C_4}{ \times ^4}{C_0} = \dfrac{{24}}{{24 \times 1}} \times \dfrac{{24}}{{1 \times 24}} \\
  { \Rightarrow ^4}{C_4}{ \times ^4}{C_0} = 1
 \]

Case 2: Now, we will select only the wives out of 4 couples for the committee. The number of ways to select 4 wives out of 4 wives and no husband is,
\[
  ^4{C_0}{ \times ^4}{C_4} = \dfrac{{4!}}{{0!\left( {4 - 0} \right)!}} \times \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}} \\
  { \Rightarrow ^4}{C_0}{ \times ^4}{C_4} = \dfrac{{4!}}{{0!4!}} \times \dfrac{{4!}}{{4!0!}} \\
  { \Rightarrow ^4}{C_0}{ \times ^4}{C_4} = \dfrac{{24}}{{1 \times 24}} \times \dfrac{{24}}{{24 \times 1}} \\
  { \Rightarrow ^4}{C_0}{ \times ^4}{C_4} = 1
 \]
Case 3: Now, we will select any 1 husband out of four husbands and then select the 3 wives such that they are not a couple. The number of ways to select 4 husbands and 3 wives are,
\[
  ^4{C_1}{ \times ^3}{C_3} = \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \times 1 \\
  { \Rightarrow ^4}{C_1}{ \times ^3}{C_3} = \dfrac{{4!}}{{1!3!}} \\
  { \Rightarrow ^4}{C_1}{ \times ^3}{C_3} = \dfrac{{24}}{{1 \times 6}} \\
  { \Rightarrow ^4}{C_1}{ \times ^3}{C_3} = 4
 \]
Case 4: Now, we will select any 2 husbands out of 4 husbands and then select the 2 wives such that they are not a couple. The number of ways to select 2 husbands and 2 wives are,
\[
  ^4{C_2}{ \times ^2}{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \times 1 \\
  { \Rightarrow ^4}{C_2}{ \times ^2}{C_2} = \dfrac{{4!}}{{2!2!}} \\
  { \Rightarrow ^4}{C_2}{ \times ^2}{C_2} = \dfrac{{24}}{{2 \times 2}} \\
  { \Rightarrow ^4}{C_2}{ \times ^2}{C_2} = 6
 \]
Case 5: Now, we will select any 3 husbands out of 4 husbands and then select the 1 wife such that they are not a couple. The number of ways to select 3 husbands and 1 wife are,
\[
  ^4{C_3}{ \times ^1}{C_1} = \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} \times 1 \\
  { \Rightarrow ^4}{C_3}{ \times ^1}{C_1} = \dfrac{{4!}}{{3!1!}} \\
  { \Rightarrow ^4}{C_3}{ \times ^1}{C_1} = \dfrac{{24}}{{6 \times 1}} \\
  { \Rightarrow ^4}{C_3}{ \times ^1}{C_1} = 4
 \]
As there are different members in the committee that can be formed for different cases, we will add the results of all the cases to get the different members for a committee that can be formed.
So, the total number of ways to form a committee is 1+1+4+6+4=16.
So, the number of different committees that can be formed in which no couple finds a place is 16.

Note: When we select the number of husbands, remember to select the number of wives out of only those husbands who are not selected to become a member of the committee.