There are 6 students participating in a badminton competition. I n the first round,they are divided into 3 pairs. The winner in each pair will proceed to the next round.
a)How many combinations are there in the first round?
b)Four more stduents , including Ada and Pan , are participating in the competition. Find the probability that Ada aginst Pan play against each other in the first round.
+118696 Okay I know why mine is wrong.
(6!2!×(6−2)!)×(4!2!×(4−2)!)=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.
+23254 I'm afraid that my answer might now have been correct, so I deleted it.
+118696 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
(6!2!×(6−2)!)×(4!2!×(4−2)!)=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. 
+33654 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:
.
+118696 Thanks Alan, I did think mine looked rather huge. Can you see where my logic is floored?
+118696 Okay I know why mine is wrong.
(6!2!×(6−2)!)×(4!2!×(4−2)!)=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.
