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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?

12 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

12 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 ?

12 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

5 years ago

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 from __5 __females (The two females excluded are the wives of the selected males). So we use 5P2 for females.

Thus total no. of arrangements of all possible matches becomes 5P2 * 7P2.

4 years ago

4 years ago

2 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})

one year ago

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

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