Problem Solving

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

answer is 840

regards,
apoorva

You have 7 men and 7 women and in mixed doubles, there has to be one man and one woman on each side.

There are 2 pairs, each with 2 partners.

For the first partner of the first pair, you have 7 choices (all of the men or all of the women).

Once you have made that choice, then you will have 6 choices left from the other gender (as one will be eliminated since that one is married to the fist choice).

Now, to the other side of the net. You have 6 men and 6 women left in the group, but one woman and one man are the spouses of the first pair, so you can't choose them, so you have only 5 men and 5 women you can choose from.

So, you have 5 choices for the first partner of the second pair (all that remain eligible from either gender).
For the second partner of the second pair, you have 5 of that gender left, but one is married to the one you just chose and so is eliminated, so you have 4 choices left.

7*6*5*4 = 840

-Carrie

Would they assume that you know what a Lawn Tennis mixed double match is?

I don't know exactly what lawn tennis is, but mixed doubles is pretty explanatory - double = 2 and mixed would be one of each.

However, I don't know if the GMAT would expect that knowledge or not.

-Carrie

well lawn tennis is something that is currently goinh on in the form of Australian Open


and mixed doubles is one male and one female as very rightly pointed out by carrie

regards,
apoorva

gaggleofgirls wrote:You have 7 men and 7 women and in mixed doubles, there has to be one man and one woman on each side.

There are 2 pairs, each with 2 partners.

For the first partner of the first pair, you have 7 choices (all of the men or all of the women).

Once you have made that choice, then you will have 6 choices left from the other gender (as one will be eliminated since that one is married to the fist choice).

Now, to the other side of the net. You have 6 men and 6 women left in the group, but one woman and one man are the spouses of the first pair, so you can't choose them, so you have only 5 men and 5 women you can choose from.

So, you have 5 choices for the first partner of the second pair (all that remain eligible from either gender).
For the second partner of the second pair, you have 5 of that gender left, but one is married to the one you just chose and so is eliminated, so you have 4 choices left.

7*6*5*4 = 840

-Carrie

Hi Carrie,

I have a query on your above explanation, Since 840 ways of selecting two mixed doubles couples will include 2! ways of rearranging these two teams so we would need divide 840 by 2! and hence the answer would be 420.

In the URL copied below, Ian Stewart explained similar problem clearly. request you to look into this and clarify.

https://www.beatthegmat.com/friends-want ... 18991.html


Thanks
Gopi

I dont think the answer can be 840. We would be over counting this.

Essentially, we are choosing 2 men and 2 women who will be playing the game.

So, we can select 2 men from seven in 7choose2 ways
=7!/(2!*5!)
= 21

Now, we need to avoid the wives of these 2 men. So we have total of 5 choices and need to select 2.

we can select 2 women from 5 in 5choose2 ways
=5!/(2!*3!)
= 10


Hence total ways for selecting 2 men and 2 women
= 21 * 10
= 210

Now that we have 2 men (m1 and m2) and 2 women (w1 and w2). We can make teams using these 4 people.

Based on the restriction of mixed doubles,
We can get a match between
(m1, w1) and (m2, w2)
or
(m1, w2) and (m2, w1)


Hence there are total 210 * 2 = 420 ways.

I think the answer must be 420.

Anyone else thinks 420 is the correct answer?? If not then could you please explain why not?? Or let me know where i went wrong. Thanks.

Appreciate the feedback.

gaggleofgirls wrote:You have 7 men and 7 women and in mixed doubles, there has to be one man and one woman on each side.

There are 2 pairs, each with 2 partners.

For the first partner of the first pair, you have 7 choices (all of the men or all of the women).

Once you have made that choice, then you will have 6 choices left from the other gender (as one will be eliminated since that one is married to the fist choice).

Now, to the other side of the net. You have 6 men and 6 women left in the group, but one woman and one man are the spouses of the first pair, so you can't choose them, so you have only 5 men and 5 women you can choose from.

So, you have 5 choices for the first partner of the second pair (all that remain eligible from either gender).
For the second partner of the second pair, you have 5 of that gender left, but one is married to the one you just chose and so is eliminated, so you have 4 choices left.

7*6*5*4 = 840

-Carrie

420 should be the answer. 7 men to start with, so 7c2 or 21

only 5 women to choose from since wives don't count so 5c2 = 10

since this states that it is a mixed double....we can only have 21*10*2 = 420.