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Math Help - Combinatorics question

  1. #1
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    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 n!/a!b!c! a=2 wins b=3 losses c= 2 ties.

    and that is where i am please help!
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  2. #2
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    Re: Combinatorics question

    Quote Originally Posted by Gurp925 View Post
    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 n!/a!b!c! a=2 wins b=3 losses c= 2 ties.
    How many ways can you arrange the string WWWL~?

    How many ways can you arrange the string WWLLTT~?

    Now multiply.
    Thanks from Gurp925
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  3. #3
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    Re: Combinatorics question

    Quote Originally Posted by Gurp925 View Post
    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 n!/a!b!c! a=2 wins b=3 losses c= 2 ties.

    and that is where i am please help!
    Try instead \frac{6!}{2! 2! 2!}.
    Thanks from Gurp925
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  4. #4
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    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.
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