Question

The number of ways in which a mixed double tennis game can be arranged from amongst 9 married couples if no husband & wife plays in the same game is?
(A) 756
(B) 3024
(C) 1512
(D) 6048

Answer 1

**** We have 9 male players and for the mixed double tennis game, we have to select 2 male players. We know that the number of ways to select m items out of n items given is $^{n}{{C}_{m}}$ . Use this formula and get the number of ways to select two husbands out of 9 male players. It is given that no husband and wife play in the same game. So, for the selection of female players for the same game we have to ignore those wives whose husbands are already selected for the game. Therefore, now, we have to select two female players out of seven female players. We know that the number of ways to select m items out of n items given is $^{n}{{C}_{m}}$ . Use this formula and get the number of ways to select two female players out of 7 female players. We know the formula, the rearrangement of x items is x!. Here, the rearrangement of female players is possible and the number of female players in a game is 2. So, the possible number of rearrangements is 2!. Now, solve it further and calculate the number of ways in which a mixed double tennis game can be arranged from amongst 9 married couples if no husband and wife play in the same game.

Complete step by step solution:

According to the question, we have 9 married couples and we have to find the number of ways in which a mixed double tennis game can be arranged from amongst 9 married couples if no husband & wife plays in the same game.

Let us understand it with a diagram,

In the above diagram, we have shown husbands with their corresponding wives.

Here, husbands are male players while the wives are female players.

We know the formula, the number of ways to select m items out of n items given = $^{n}{{C}_{m}}$ …………………..(1)

For the mixed double tennis game, we need two male players out of 9 male players.

Using the formula shown in equation (1), we get

The number of ways to select two husbands out of 9 male players = $^{9}{{C}_{2}}$ .

It is given that no husband and wife play in the same game. So, for the selection of female players for the same game we have to ignore those wives whose husbands are already selected for the game.

Suppose we have selected the male players, 1 H and 2 H. So, we have to ignore the wife of 1 H and 2 H. It means for the male players 1 H and 2 H, we have to select the two female players out of seven female players that are 3 W, 4 W, 5 W, 6 W, 7 W, 8 W, and 9 W.

Using the formula shown in equation (1), we get

The number of ways to select two wives out of 7 wives = $^{7}{{C}_{2}}$ .

We know the formula, the rearrangement of x items is x!.

Here, the rearrangement of female players is also possible and the number of female players in a game is 2. So, the possible number of rearrangements is 2!.

In the mixed double tennis game, we need male players as well as female players.

So, the total number of ways = $^{9}{{C}_{2}}{{\times }^{7}}{{C}_{2}}\times 2!=\dfrac{9\times 8}{2}\times \dfrac{7\times 6}{2}\times 2=9\times 4\times 7\times 6=1512$ .

Hence, the correct option is (C).

Note:**** In this question, one might think that rearrangement of female players is not possible.

Suppose we select the male players, 3 H and 4 H. It is given that no husband and wife play in the same game. So, we have to ignore the wife of 3 H and 4 H. It means for the male players 3 H and 4 H, we have to select the two female players out of seven female players that are 1 W, 2 W, 5 W, 6 W, 7 W, 8 W, and 9 W. For instance, we select W 5 and W 6 as our female players. Now, for the game, we have two teams. The first team can be of 3 H and W 5, and the second team can be of 4 H and W 6. There is also a possibility that the first team can be of 3 H and W 6, and the second team can be of 4 H and W 5. Now, we can observe that the rearrangement of female players is possible here.