Question

Suppose that A and B are playing a tennis match in which the first player to win three sets out of 5 sets wins the match. In how many ways could this match end?
Please, help

Answer 1

This what I did:
To win, A must be the last team win the last match. So that the first four sets will be AABB
and we have \(\dfrac{4!}{2!2!}=6\)ways to have A win.
The same for B win.
So that we have 12 ways to have the match end.

Answer 2

However, my prof's solution is 20. I don't know why.

Answer 3

2(1+5c3)

Answer 4

He said, we have to consider the cases:
AAA and AABA (which is 3 possibilities)

Answer 5

sorry answer is 2(5c3)

Answer 6

However, if AAA, then the match doesn't last to 5sets to end.

Answer 7

my head can't work hehe. good luck friend^_^

Answer 8

lets say A wins.. so A needs 3 wins out of 5 (x x x x x ) we just need to arrange 3 wins in these 5 crosses..

Answer 9

This is my test. I failed this problem but didn't pass the bad feeling yet. HIHIHIHI

Answer 10

Nope, some students argued \(\dfrac{5!}{2!3!}\) = 10 and times2 to get 20. The right answer but the wrong method. Still get 0.

Answer 11

if A wins after 3 sets, that means A has hatrick, AAA, and only 1 case for this happen
at the end, the possibility for A to win is 1 + 3+6 =10
Damn!!!

Answer 12

Thanks for reply. I got it now. I deserved bad credit. hihihi... satisfy now.

Answer 13

take my medal back ..lol.. my method wasnt correct .