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A mixed doubles tennis game is to be played between two teams (each team consists of one male and one female). There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

GIVEN ANS 42 ,

i am getting 36..my approach was as given team has no husband and his wife so i can write like there are 4 husband and no wife at their correct husband (dearrangement=n ) these are no of teams and total no of games will be nc2 .

where am i missing ?
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There are $4$ married couples & No husband can be team up with his wife.

Assuming, $P, Q, R, S$ are the husbands & $A,B,C,D$ are the wives.

So, the married couples are : $(P,A);(Q,B);(R,C);(S,D)$

Now, we'll choose $2$ husbands first from the $4$ husbands i.e. $^4C_2 = 6$

∴ There are $6$ ways we can choose Males for the games.

Now, If we choose $2$ male think $P$ & $Q$, then we'll get $2$ cases:

CASE 1 : 

  • If $C$ or $D$ is assign with $P$. This can be done in only $2$ ways
  • Now, there will be $2$ choices or $2$ female available to be coupled with $Q$.( i.e either $C$ or $D$ and $A$) This can be done in $^2C_1 \text{ ways}$

Total no. of ways of CASE 1: $^4C_2 \times 2 \times ^2C_1 = 6 \times 2 \times 2 = 24$

CASE 2:

  • If $B$ is coupled with $P$, can be done in $1$ ways only.
  • We've $3$ choices (females ) left with to couple with $Q$ & this can be done in $^3C_1 \hspace{0.1cm} ways$

Total no. of ways of CASE 2: $^4C_2 \times 1 \times ^3C_1 = 6 \times 1 \times 3 = 18$

∴$\color{Green}{\text{ Maximum no. of games that can be played is}}$ = $\text{CASE 1  OR CASE 2}$

$\qquad \qquad \qquad \qquad = \color{teal}{24 + 18 \hspace{0.1cm} ways = 42 \hspace{0.1cm} ways}$

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Given there are 2 teams and each team has 2 players

So, First pick 1 team

Case 1:

Take 2 male among 4 in $\binom{4}{2}$ ways(Say $M_{1}$)

Case 2:

Now pick female

Here are 2 possibilities

1)if  $M_{1}$ already choose $M_{2}$'s wife , Among rest 3 female $M_{2}$ can choose $M_{1}$ in $\binom{1}{1}$ ways

2)  If $M_{1}$ not choosen $M_{2}$'s wife , Among rest 2 female $M_{2}$ can choose any one in $\binom{2}{1}$ ways and ignore his own wife

and they can arrange in $2!$ ways

So, Total possibility $6 \times 2=12$ ways
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