# In how many ways can a lawn tennis mixed doubles be made up from seven married couples if no husband and wife play in the same set?

In how many ways can a lawn tennis mixed doubles be made up from seven married couples if no husband and wife play in the same set?

## 10 Answers

SELECT 2 PERSONS FROM 14

NO. OF WAYS = 14C2=91

IN THISTHERE ARE 7 WAYS IN WHICH HUSBAND WIFE COME TOGETHER

TOTAL WAYS= 91-7=84

OUT OF THESE 84 TEAMS WE HAVE TO SELECT 2 TEAMS

NO. OF WAYS = 84C2 = 3486 WAYS ANS

Thanks for replying. But the answer is 840 only. I think u misjudged the question. in first step u took 14C2 , in this case it is possible that we select 2 husbands but no wife... instead of this it must be like this

7C2 x 7C2 since we want two husbands but two wives also. these are 21 x 21 = 441 ways.

now we want to deduct the unfavorable cases i.e 14 cases i m stuck up at this point how to get the answer ahead?

can anyone help?

actually this question is given in permutation chapter can this be solved without combination ?

it should be 210.

as u select two husbands u will have five wives left to select from.......

a regular polygon of 9 sides each of length 2 unit is inscribed in a circle the radius of circle is

**females (The two females excluded are the wives of the selected males). So we use 5P2 for females.**

__5__^{7}c

_{2}*

^{5}c

_{2})

Please find below the solution to your problem.

Let the two husbands A, B be selected out of seven males in 7C2 = 21 ways. Excluding their wives we have to select two ladies C, D out of remaining 5 wives in 5C2 = 10 ways.

Thus the number of ways of selecting the players for mixed double is 21 x 10 = 210.

Now suppose A chooses C as a partner (B will automatically go to D) or A chooses D as a partner, (B will automatically go to C). Thus we have 2 choices for the teams. Required number of ways is 210 x 2 = 420.

Thanks and Regards