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1.Find the number of ways in which a mixed double tennis game can be arranged between 10 players consisting of 6 men and 4 women.

My approach : Since,if every pair of selection of men and women gives rise to 4 matches so C(6,2) * C(4,2) * 4.

2.Find the number of ways in which a mixed double game can be arranged from amongst nine married couples, if no husband and wife play in the same game.

My approach : C(9,2) * C(7,2) * 4.

Now, this two questions comes from my test paper I would be grateful if somebody check whether this is correct or not since my answers doesn't matched to the given solution in the paper.

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I get a final factor of 2 rather than 4. –  Robin Chapman Oct 4 '10 at 9:08
    
@ Robin Chapman : Could you please explain your idea ? My idea is like this if we select M1,M2 and W1,W2 then the team can be constituted in 4 ways : M1,W1; M2,W2; M1,W2; M2,W1; –  Quixotic Oct 4 '10 at 9:20
    
Debanjan, that looks like two possible matches to me. –  Robin Chapman Oct 4 '10 at 9:56
    
Sorry i didn't get you :| –  Quixotic Oct 4 '10 at 10:25
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@ Debanjan: I think your list is the list of possible teams, not the list of possible matches (M1,W1)v(M2,W2), (M1,W2)v(M2,W1). –  tttppp Oct 4 '10 at 11:45

1 Answer 1

up vote 2 down vote accepted

The answer to (1) is 2*C(6,2)*C(4,2) (and not with a 4), since M1, W1 playing M2, W2 is the same as M2, W2 playing M1, W1. I don't think the question wants you to take account of which end they're playing from. You've made a similar error in part(2).

To make it more clear you've included a factor 2 for swapping the women over, say, in our set of four people, and then included another factor of 2 for swapping the men over, but have of course ended up with the same match when both swaps have taken place.

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So in both the problem the factor will be 2 ? –  Quixotic Oct 4 '10 at 18:54
    
@Debanjan: Yes. –  Derek Jennings Oct 4 '10 at 20:52

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