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
