Question

Number of ways in which a lawn-tennis mixed double be made from seven married couples if no husband and wife play in the same set is
(a) 240
(b) 420
(c) 720
(d) none of these

Answer 1

Hint: We will solve this question with the help of permutations and hence we will use the formula \[{}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\]. We will first arrange 2 men out of 7 men and then we will arrange 2 women out of 5 women as these two women should not be with their respective husbands.

Complete step-by-step answer:
Now it is mentioned in the question that the number of men is 7 and also the number of women is 7.
Number of ways to arrange 2 men out of 7 men \[={}^{7}{{P}_{2}}.........(1)\]
It is given that no husband and wife play in the same set, then number of ways to arrange 2 women out of 5 women (Wives of the two men cannot play in same game) \[={}^{5}{{P}_{2}}.........(2)\]
Hence from equation (1) and equation (3) and by the fundamental principle of counting, the required number of ways \[={}^{7}{{P}_{2}}\times {}^{5}{{P}_{2}}.........(3)\]
Now with the help of the formula \[{}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\] in equation (3) we get,
\[\Rightarrow \dfrac{7!}{(7-2)!}\times \dfrac{5!}{(5-2)!}........(4)\]
Now expanding the factorials in equation (4) we get,
\[\begin{align}
  & \Rightarrow \dfrac{7!}{5!}\times \dfrac{5!}{3!} \\
 & \Rightarrow \dfrac{7\times 6\times 5!}{5!}\times \dfrac{5\times 4\times 3!}{3!}.........(5) \\
\end{align}\]
Now cancelling similar terms in equation (5) and solving we get,
\[\Rightarrow 42\times 20=840\]
Hence the number of ways is 840. So the correct answer is option (d).

Note: Permutation relates to the act of arranging all the members of a set into some sequence or order. And knowing this formula \[{}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\] is important . We can make a mistake in solving equation (4) if we do not properly expand \[{}^{7}{{P}_{2}}\] and \[{}^{5}{{P}_{2}}\].