Question

A man has 7 relatives, 4 of them are ladies and 3 gentlemen; his wife has 7 relatives in which 3 of them are ladies and 4 are gentlemen. The number of ways they can invite a party of 3 ladies and 3 gentlemen so that there are 3 man’s relatives and 3 of wife’s relatives is:
(a) 455
(b) 565
(c) 485
(d) None of these

Answer 1

Hint: We have to select 3 relatives from man’s and 3 relatives from his wife’s side to invite a party for 3 ladies and 3 gentlemen. We are using combinatorial approach to solve this problem n which we have to make cases like first case is selecting 3 gentlemen from man’s relatives and selecting 3 ladies from his wife’s relatives second case would be 2 gentlemen and 1 lady from man’s relatives and selecting 2 ladies and 1 gentlemen from his wife’s relatives likewise make all the cases and add all the cases to get the total number of ways.

Complete step-by-step answer:
We have to find the number of ways to select 3 ladies and 3 gentlemen out of the 7 relatives of men and 7 relatives of his wife in such a way that 3 man’s relatives and 3 his wife’s relatives should be involved.
It is also given that a man has 7 relatives, 4 of them are ladies and 3 gentlemen and also given that his wife has 7 relatives in which 3 of them are ladies and 4 are gentlemen
Case (1):
When we are selecting 3 gentlemen from man’s relatives and 3 ladies from his wife’s relatives:
For that to happen we are using a combinatorial way by selecting 3 gentlemen out of 3 gentlemen from man’s side and selecting 3 ladies out of 3 ladies from his wife’s relatives then multiplying both the selections.
${}^{3}{{C}_{3}}\left( {}^{3}{{C}_{3}} \right)$
We know that ${}^{n}{{C}_{n}}=1$ so using this relation in the above expression we get,
$\begin{align}
  & 1(1) \\
 & =1 \\
\end{align}$
Case (2):
When we are selecting 2 gentlemen and 1 lady from man’s relatives and 1 gentlemen and 2 ladies from his wife’s relatives:
Selecting 2 gentlemen out of 3 gentlemen and selecting 1 lady out of 4 ladies from the man’s relatives and selecting 1 gentleman out of 4 gentlemen and 2 ladies out of 3 ladies from his wife’s side and then multiply all of them.
${}^{3}{{C}_{2}}\left( {}^{4}{{C}_{1}} \right)\left( {}^{4}{{C}_{1}} \right)\left( {}^{3}{{C}_{2}} \right)$
We can write ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ so using this relation in the above expression we get,
${}^{3}{{C}_{1}}\left( {}^{4}{{C}_{1}} \right)\left( {}^{4}{{C}_{1}} \right)\left( {}^{3}{{C}_{1}} \right)$
We know that ${}^{n}{{C}_{1}}=n$ and using this relation in the above equation we get,
$\begin{align}
  & 3\left( 4 \right)\left( 4 \right)3 \\
 & =9\left( 16 \right) \\
 & =144 \\
\end{align}$
Case (3):
When we are selecting 1 gentleman and 2 ladies from man’s relatives and 2 gentlemen and 1 lady from his wife’s relatives:
Selecting 1 gentleman out of 3 gentlemen and selecting 2 ladies out of 4 ladies from the man’s relatives and selecting 2 gentleman out of 4 gentlemen and 1 lady out of 3 ladies from his wife’s side and then multiply all of them.
${}^{3}{{C}_{1}}\left( {}^{4}{{C}_{2}} \right)\left( {}^{4}{{C}_{2}} \right){}^{3}{{C}_{1}}$
We know that:
$\begin{align}
  & {}^{4}{{C}_{2}}=\dfrac{4!}{2!2!} \\
 & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4.3.2!}{2!2!} \\
\end{align}$
$\begin{align}
  & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{12}{2.1} \\
 & \Rightarrow {}^{4}{{C}_{2}}=6 \\
\end{align}$
Substituting the value of ${}^{4}{{C}_{2}}=6$ in the above expression we get,
$\begin{align}
  & 3\left( 6 \right)\left( 6 \right)3 \\
 & =9\left( 36 \right) \\
 & =324 \\
\end{align}$
Case (4):
When we are selecting 3 ladies from man’s relatives and 3 gentlemen from his wife’s relatives:
For that to happen we are using a combinatorial way by selecting 3 ladies out of 4 ladies from man’s side and selecting 3 gentlemen out of 4 gentlemen from his wife’s relatives then multiplying both the selections.
${}^{4}{{C}_{3}}\left( {}^{4}{{C}_{3}} \right)$
We can write ${}^{4}{{C}_{3}}={}^{4}{{C}_{1}}$ and substitute this value in the above expression we get,
$\begin{align}
  & {}^{4}{{C}_{1}}\left( {}^{4}{{C}_{1}} \right) \\
 & =4\left( 4 \right) \\
 & =16 \\
\end{align}$
Now, to get the total number of ways we are adding all the three cases which will give us:
$\begin{align}
  & 1+144+324+16 \\
 & =485 \\
\end{align}$
Hence, the correct option is (c).

Note: You might have been thinking why we have multiplied the combinatorial ways in each case. For instance in case (1):
${}^{3}{{C}_{3}}\left( {}^{3}{{C}_{3}} \right)$
 Because selecting 3 numbers out of 3 ladies is independent of selecting 3 gentlemen out of 3 gentlemen and we know that when events are independent to each other then we put a multiplication sign.
Similarly, we have added all the cases because each case is fulfilling the condition that selecting 3 ladies and 3 gentlemen out of the 7 relatives of men and 7 relatives of his wife in such a way that 3 man’s relatives and 3 his wife’s relatives should be involved.