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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?`
8 years ago

ARCHIT SINGHAL
18 Points
```										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
```
8 years ago
yudhisther sharma sharma
18 Points
```										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 ?
```
8 years ago
nishant kumar agarwal
20 Points
```										it should be 210.
as u select two husbands u will have five wives left to select from.......
```
7 years ago
praman gupta
16 Points
```										a regular polygon of 9 sides each of length 2 unit is inscribed in a circle the radius of circle is
```
6 years ago
11 Points
```										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
```
one year ago
Prateek Garg
26 Points
```										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.
```
9 months ago
DG
11 Points
```										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 months ago
Saurabh Jain
19 Points
```										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.From Vidyasagar Academy East Delhi 9910791967
```
3 months ago
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