# Thread: Combinatorics question

1. ## Combinatorics question

Hello!, My question is A hockey team has played 10 games and has a record of 5 wins, 3 losses and 2 ties. In how many ways could this have happened if after the first 4 games the team's record was 3 wins and a loss.

This is what i thought to do

1)_ _ _ _ _ _ _ _ _ _ first 4 were W-W-W-L Leaving us to deal with 6 more

2) i asked myself what are the common things here, 2 wins 3 losses and 2 ties and an n value of 6 remaining.

3) so i popped it into the formula $\displaystyle n!/a!b!c!$ a=2 wins b=3 losses c= 2 ties.

and that is where i am please help!

2. ## Re: Combinatorics question

Originally Posted by Gurp925
Hello!, My question is A hockey team has played 10 games and has a record of 5 wins, 3 losses and 2 ties. In how many ways could this have happened if after the first 4 games the team's record was 3 wins and a loss.
This is what i thought to do
1)_ _ _ _ _ _ _ _ _ _ first 4 were W-W-W-L Leaving us to deal with 6 more
2) i asked myself what are the common things here, 2 wins 3 losses and 2 ties and an n value of 6 remaining.
3) so i popped it into the formula $\displaystyle n!/a!b!c!$ a=2 wins b=3 losses c= 2 ties.
How many ways can you arrange the string $\displaystyle WWWL~?$

How many ways can you arrange the string $\displaystyle WWLLTT~?$

Now multiply.

3. ## Re: Combinatorics question

Originally Posted by Gurp925
Hello!, My question is A hockey team has played 10 games and has a record of 5 wins, 3 losses and 2 ties. In how many ways could this have happened if after the first 4 games the team's record was 3 wins and a loss.

This is what i thought to do

1)_ _ _ _ _ _ _ _ _ _ first 4 were W-W-W-L Leaving us to deal with 6 more

2) i asked myself what are the common things here, 2 wins 3 losses and 2 ties and an n value of 6 remaining.
This is incorrect. You initially said there were 3 losses overall and they had 1 loss in the first four games. That leaves 2 wins, 2 losses, and 2 ties in the remaining 6 games.

3) so i popped it into the formula $\displaystyle n!/a!b!c!$ a=2 wins b=3 losses c= 2 ties.

and that is where i am please help!
Try instead $\displaystyle \frac{6!}{2! 2! 2!}$.

4. ## Re: Combinatorics question

What an oversight my mistake yeah that is right in the last 6 games there are 2 wins 2 losses and 2 ties, * not 3 as i miscounted* great that gives me 90 which i think is the answer thank you both,

Hallsofivy and Plato you guys are champs appreciate you always helping out.