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.

Guest Feb 2, 2015

#5**+5 **

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.

Melody
Feb 3, 2015

#2**+5 **

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.

Melody
Feb 2, 2015

#3**+5 **

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:

.

Alan
Feb 2, 2015

#4**0 **

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

Melody
Feb 2, 2015

#5**+5 **

Best Answer

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.

Melody
Feb 3, 2015