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First term gives us number of ways for team to win 3 of its total of 6 games. Second term tells us how many ways there are for the team to lose 2 of its remaining 3 games. Last term is of course 1C1, as there is only one game left.

why do we permute across 6 slots and account for the repeats? i dont understand the logic of it.

Thanks for making me rethink ....I thought on different lines...this is not the way this ques should be solved.

I think the correct approach is: 6C3 *3C2 *1C1 = 60

I think no. of ways = 6!/3!2! = 60 is correct.

6!=6P6 - all possible permutation with 3 wins, 2 losses and 1 draw 3!=3P3 - we exclude options for wins: (w1,w2,w3,l,l,d) and (w3,w1,w2,l,l,d) are the same way. 2!=2P2 - we exclude options for losses: (w,w,w,l1,l2,d) and (w,w,w,l2,l1,d) are the same way.
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To eliminate repetitions...shouldn't you be subtracting them from 6!......if at all you choose to do it that way. Why would you divide...can you explain?

Also, doing permutation and then eliminating repetitions isn't effectively a combination.

To eliminate repetitions...shouldn't you be subtracting them from 6!......if at all you choose to do it that way. Why would you divide...can you explain?

We have two independent options. For example, three-digits integer: total number of options is 999-100+1=900 the number of options for the first (units) digit: 10 the number of options for the second and third digits: 900/10=90 (division)

Moreover, nCm=nPm/mPm - the same situation with elimination of repetitions

kyatin wrote:

Also, doing permutation and then eliminating repetitions isn't effectively a combination.

For me it is an effective under-1-min way. In the case of permutation-combination problems I prefer to have a few methods
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in case of repeated elements ([b]we have 3 wins and 2 losses)[/b] we must divide the total number! by the repeated elements!. in this case 6!/3!*2!. this is a formula