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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?


i am getting 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 ?
in Numerical Ability by Boss (17.1k points)
edited by | 266 views
I'm getting 42
check my answer, i think it'll clear the doubt

The value of dearrangement is 9 rt.

But after that whyare you calculating 9C2.

Dearrangement gives the number of ways set of teams can be formed satisfying the above condition.

For example if males are (M1,M2,M3,M4) and women are (W1,W2,W3,W4)

Then 1 way to form a set of  teams is {(M1,W2),(M2,W3),(M3,W4),(M4,W1)}

Another way to form set of  teams is {(M1,W3),(M2,W4),(M3,W1),(M4,W2)}

In this way we can have 9 sets of teams.
sorry given ans is 42 only ... whats thee mistake in my method ...

they are asking total noo of games ..thats why 9c2
check my approach

by derangement how getting 36?

write ur calculation
by dearrangement am getting 9 .. then 9c2 =36 ,

$\sum_{r=2}^{4}(-1)^r n!/r!$
@Sonam value of dearrangement is not equal to the number of teams.
ok why ?
Here dearrangement gives the value of how many ways we can form 4 mixed doubles teams such that no husband gets his wife as his partner.


is the case where each player handshake with each other player

got it?

then how derangement will happen here?

 Kushagra Chatterjee, may i know why u're giving downvote, am i doing wrong?

if u don't understand anyone's approach, that doesn't mean he/she is doing wrong

And yes, you get the downvoting capability in case if u face anything spam, or inappropriate respective to that questions

do u mean

 in 2nd team female should be choosen  like that(among 3 if B's own wife in team 1 or among 2 if B's own wife is not still selected)
srestha team does not contain husband with his wife thats why i did dearrange ..

where u have done derangement in calculation?

u have done simple selection

$\binom{9}{2}$ =36

where derangement here?

@Sonam @Soumya @Subarna

12 will be ans

chk here

why 42 ways needed , 12 can simply solve the problem


sresth this is arun sharma solved example given ans is 42 only

2 Answers

+3 votes
Best answer

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: $^4C_2 \times 2 \times ^2C_1 = 6 \times 2 \times 2 = 24$


  • 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: $^4C_2 \times 1 \times ^3C_1 = 6 \times 1 \times 3 = 18$

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

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

by Boss (17.8k points)
selected by
In case 1 bullet number 2

You telling there are 3C1 ways but at the time of multiplying you are multiplying with 2C1.
is the ans not be 54?

 Kushagra, just a typo :P

no, i think it'll be $42$
Yes, It should be $42$.

Number of ways to form a team = $4*3 = 12$ ways
(There are 4 husbands and each of them has 3 choices to pair with).

Now out of these 12 teams- each team can have a match with 7 teams only.
Reason - Suppose $(A,Q)$ is a team. Now I have to subtract those possible teams which include A or Q.

For Example - $(A,Q)$ can't play with $(A,Q),(A,R),(A,S),(C,Q),(D,Q)$.

Number of matches possible = $\Large\frac{12*7}{2}=42$
Subarna you are right may be I am not getting your answer. I am extremely sorry I have downvoted it. I am removing it.

In case 1

I think you are trying to count the number ways teams can be formed when P is paired with C or D such that no husband gets his wife as a partner. Please tell me whether I am right or wrong.

If I am right then the team (R,C) is also counted.

When P is paired with D

           Q is paired with A

           R is paired with C

           S is paired with B
I'm assuming (P,A);(Q,B);(R,C);(S,D) - this is the original Husband-Wife pair.
Then, i'm choosing 2 males first from 4 males.

so, i have $^4C_2 = 6$ possible ways to find two male
I'm taking P,Q as selected males

But, there can be another $5$ ways to select the males.

they are: (P,R) OR (P,S) OR (Q,R) OR (Q,S) OR (R,S)
Yes I am telling that since R and C are original couple R can't team up with C but in your counting R can team up with C.
No, R is not teamed up with C
How u're getting R,C through my approach?
So,P and Q are the selected males

Assume P is teamed up with D

                Q is teamed up with A

       And  S is teamed up with B

        Then R is teamed up with C
Female partners are going to change according to your chosen male group
I am not getting it can u please explain it a bit more.
when P and Q playing the game R is out of game and cannot pair up with anyone.

So, R,C pair canceled out at that time
Ok now I got the point. I get confused sometimes.

@Somya29 Very well explained!!
0 votes
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
by Veteran (119k points)
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