Question

In how many ways can a mixed double game can be arranged amongst eight married couples if no husband and wife play in the same game.
A. \[840\]
B. \[460\]
C. \[580\]
D. None of above

Answer 1

Hint:Here we use the knowledge of a mixed double game which consists of two teams of one husband and one wife each and using the method of combination we find ways to choose two husbands from total number of men and then to choose wives we subtract those two wives whose husbands are already chosen for the game.
* Combination formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\] where factorial opens up as \[n! = n \times (n - 1) \times (n - 2)......3 \times 2 \times 1\]

Complete step-by-step answer:
Total number of couples is \[8\] which means there are \[8\] husbands and \[8\] wives.
For a mixed doubles game there are two teams of one couple each but here there is a requirement that no husband plays with their wife in one team.
So first we find number of ways to choose two husbands from total number of husbands.
Since \[n = 8,r = 2\] here so substitute in formula for combination \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
\[^8{C_2} = \dfrac{{8!}}{{(8 - 2)!2!}}\]
        \[
   = \dfrac{{8!}}{{(6)!2!}} \\
   = \dfrac{{8 \times 7 \times 6!}}{{6!2!}} \\
 \] { since \[n! = n \times (n - 1)!\]}
Now cancel out same terms from numerator and denominator
      \[
   = \dfrac{{8 \times 7}}{2} \\
   = 4 \times 7 = 28 \\
 \]
Therefore, number of ways to choose two husbands from eight husbands is \[28\]
We have total number of wives \[8\]
Since, two husbands are selected, therefore we cannot have their wives participating in the selection process. So we subtract two wives from total number of wives.
Now, number of wives we can choose from \[ = 8 - 2 = 6\]
Therefore, now we find ways to choose two wives from six wives.
Since \[n = 6,r = 2\] here so substitute in formula for combination \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
\[^6{C_2} = \dfrac{{6!}}{{(6 - 2)!2!}}\]
        \[
   = \dfrac{{6!}}{{(4)!2!}} \\
   = \dfrac{{6 \times 5 \times 4!}}{{4!2!}} \\
 \] { since \[n! = n \times (n - 1)!\]}
Now cancel out same terms from numerator and denominator
      \[
   = \dfrac{{6 \times 5}}{2} \\
   = 3 \times 5 = 15 \\
 \]
Therefore, number of ways to choose two wives from six wives is \[15\]
So number of ways to choose teams having no husband wife playing together in the same game is
\[^8{C_2}{ \times ^6}{C_2} = 28 \times 15 = 420\]
But since, in the game the team can be switched i.e. any wife can play with any husband so there are two possibilities inside the game as well.
Therefore, total number of ways to arrange the game is \[2 \times 420 = 840\]
Thus, option A is correct.

Note:
Students are likely to make the mistake of forgetting that there are possibilities inside the game as well i.e. after choosing two husbands and two wives, we can switch them to form two teams.