Combinations - Please help!!

C

carebear

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Aug 30, 2010
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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?

I tried:

5C3 x 3C1 X 2C0 = 30

Then not sure what to do?

5C2 x 3C2 x 2C2 = 30

30 x 30 = 900.....but answer is supposed to be 360....please help me understand what I am doing wrong?
 
carebear said:
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?

I tried:

5C3 x 3C1 X 2C0 = 30

Then not sure what to do?

5C2 x 3C2 x 2C2 = 30

30 x 30 = 900.....but answer is supposed to be 360....please help me understand what I am doing wrong?
Click to expand...

What you did tells me that you try to fit in formulae somehow wherever you can. This approach will never help you in solving good questions of combinatorics. Each problem needs to be thought of individually, there is no universally applicable formula. Analyze in each problem what are the constraints and proceed accordingly.
Now, coming to the present problem. We are told there were 3 wins and 1 loss in first 4 games. No. of ways this can happen is, [4!/3!] ... 4! being the total number of permutations of 4 objects taken all at a time, and divide by 3! because 3 wins are identical. Then, we have 2 wins, 2 loses, and two ties in the last 6 games. This can happen, similar to previous case, in 6!/(2! * 2! * 2!). Multiply the two results as any one arrangement depends on both of them. You will get your answer.
 
Novice said:
What you did tells me that you try to fit in formulae somehow wherever you can. This approach will never help you in solving good questions of combinatorics. Each problem needs to be thought of individually, there is no universally applicable formula. Analyze in each problem what are the constraints and proceed accordingly.
Now, coming to the present problem. We are told there were 3 wins and 1 loss in first 4 games. No. of ways this can happen is, [4!/3!] ... 4! being the total number of permutations of 4 objects taken all at a time, and divide by 3! because 3 wins are identical. Then, we have 2 wins, 2 loses, and two ties in the last 6 games. This can happen, similar to previous case, in 6!/(2! * 2! * 2!). Multiply the two results as any one arrangement depends on both of them. You will get your answer.
Click to expand...

Thank you....understand.
 
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