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## 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?

11 years ago

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

11 years ago

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 ?

11 years ago

it should be 210.

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

10 years ago

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

9 years ago

Ans= 840 Since in a mixed doubles 2 player play in the que. One husband and one wife will play. No. Of ways husband`s can be chosed is 7p2 and no. Of ways wife could be chosen will be 5p2 as the wife of two husband chosen shouldn`t include. thus and 7p2 × 5p2= 840

4 years ago

In any of the matches, the total players playing in the game would be 4 i.e (one male one female) vs (one male one female). Also the male and female constituting the team should not be married couples. So for a match, 2 males and 2 females are essential. Therefore we select 2 males from total 7 males by 7P2. Now, females to be selected shouldn't be the wives of the 2 males selected. Therefore we select 2 females fromfemales (The two females excluded are the wives of the selected males). So we use 5P2 for females.5Thus total no. of arrangements of all possible matches becomes 5P2 * 7P2.

3 years ago

doublea match 4 players play out of which 2 are male and 2 are female.CASE 1[when males are selected first]we can select 2 males out of 7 which is 7C2Now since the two men have already been selected their wifes cannot be a part of the match hence available options are 7-2 = 5 therefore we select 2 women out of 5 i.e 5C2Total ways = 7C2 * 5C2 = 210CASE 2[ when females are selected first]now we do exactly the opp.We select 2 women from 7 and 2 men from 5 Total ways = 7C2 * 5C2 = 210GRAND TOTAL = 210 +210 = 420 AS PER MY LOGIC HOWEVER ANSWER AS PER RD SHARMA IS 840

3 years ago

If we consider the answer is 840. Which in my view is wrong. Because in 840 we have arranged both men and both women. But in doing so we have counted every possible way twice. For example WE HaveM1 M2 W1 W2Now M1 W1 vs M2 W2And M1W2 vs M2 W1 are the only possiblities. So ans must be 420.

3 years ago

the ans is 840 because according to the ans above when 2 husbands are selected from the remaining 7,wife are selected from 5 to neglect ther iwn wifes.the relatin also goes vice versa and so the ans becomes 2(^{7}c_{2}*^{5}c_{2})

11 months ago

Dear Student,

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

2 months ago

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