probability

Okay I know why mine is wrong.

$${\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{90}}$$

This does give the number of choices if the order they are chosen matters but it does not

so say  the partners are AB  CD  EF

so you can choose which of these pairs go first the is 3 choices then second 2 choices then 3rd only 1 choice,

that is 3*2=6 ways so

90 must be divided by 6 = 15 possible pairs.

I'm afraid that my answer might now have been correct, so I deleted it.

a)  Choose 2 from 6 to be the first pair that is 6C2 =15

then choose 2 from 4 for the second pair   4C2 = 6

So I think that the answer for part a is      6C2*4C2 

$${\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{90}}$$

There are 90 pair combinations.      

Now I don't understand the question for part b.  Do the 4 new ones partner each other of do they get thrown in before the original 6 get paired off?     I'm really confused now.    

For part a) I think there are only 15 sets of 3 pairs.  If the players are numbered 1 to 6 then the sets are as follows:

.

Thanks Alan, I did think mine looked rather huge.  Can you see where my logic is floored?