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+1 vote

According to my understanding, there should be men and women both in the team. So we can do:
3M and 1W or
2M and 2W or
1M and 3W.

So it will be: C(5,3)*C(5,1)+C(5,2)*C(5,2)+C(5,1)*C(5,3).

But the answer given is 600. How is it possible?

asked in Probability by Loyal (4.2k points)
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2 Answers

+1 vote
Best answer
A mixed pair in a tennis match consists of 1 man and 1 woman. To form the team you have to select 4 pairs of 1 man and 1 woman.So there'll always be 4 men and 4 women in team. Hence your approach is incorrect.

So suppose you have 5 men: M1, M2, M3, M4, M5 and 5 women: W1, W2, W3, W4, W5

Now, for first pair if you select a man, you have 5 options for women to form a team.

For second pair, if you select a man, you have 4 options for women (one women already formed the team, so she can't be included in any other team)

Similarly, for 3rd pair, you have 3 options and 4th pair you have 2 options.

Now from 5 men, you have to select 4 [ can select M1, M2, M3, M4 or M2, M3, M4, M5 or ... ] i.e., $5 \choose 4$

So total possible number of selections:  $5 \choose 4$ * 5 * 4 * 3 * 2 = 5 * 120 = 600
answered by Loyal (4k points)
selected by
+1 vote

Here, they are asking In how many ways the pairing can be done?

First, select 4 men and 4 women. Thaat can be done $\binom{5}{4}$ * $\binom{5}{4}$

Now, once selected, they can be arranged in 4! ways (i.e. onto functions from 4 men to 4 women)

So, asnwer = $\binom{5}{4}$ * $\binom{5}{4}$ * 4! ways

                 = 600 ways

answered by Veteran (18k points)

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