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not to be harsh, but almost every step here has a significant error in it... not trying to be rude but the guy who asked the question shouldn't use this as an example.

to choose all possible combinations you don't take 12x11x10x9. This is counting permutations, not combinations. division by 2 does not compensate for this, to get combinations you have to divide by

4! = 4x3x2x1 = 24,

which is the number of ways of arranging 4 unique objects among 4 people (not 16 as was stated). This division divides out the number of extra ways we would be able to arrange the prizes if they were unique. So, the number of ways of arranging the four identical prizes between four couples is

12(choose)4 = 12x11x10x9 / 4x3x2x1 = 495.

Now that we have counted combinations properly, no other multiplication is necessary to deal with selecting the couples. In the post above, the division by 2 to get combinations should be a division by 24, and there should be no multiplication after. Such a multiplication would be to account for 4 unique prizes, which is not what we have here (and, 16 is simply wrong, and, if there were unique prizes, the best advice is to count permutations, not combinations, in the first step to avoid other possible errors such as what happened there with an incorrect calculation of the number of ways to distribute 4 items among 4 people).

Now that the couples have been selected, there are two members of each couple. So for each of the 4 couples who get a prize, there are

2(choose)1 = 2

ways of them being awarded the prize. This factor multiplies what we had before, so the final result should be

12(choose)4 x 2(choose)1 = 495 x 2 = 990 ways of awarding the prizes.

As a side note, the example given in the post above using 4 people happens to work, but it is just a conicidence, there is no mathematically correct reasoning there. You do not divide by 2, you divide by r! where r is the number of identical prizes to be distributed. In the example 2 prizes were chosen, and since 2! = 2 it happened to work. If you look at the expressions for the permute and combine functions, you see that

n(choose)r = n! / r!(n-r)!

has an extra r! in the denominator when compared to

n(permute)r = n! / (n-r)!,

this is the r! i just mentioned (which accounts for the difference between unique and idential items being distributed).

yep you're right... well done and be careful what you read on these boards!

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Thank you very much for your previous help!

I have two more problems that will pretty much sum up all that i need to know.

If you can explain how to simplify these that would be GREAT!!

(-3√3)(-4√5)(2√12)

and

3y/y^2+5y-14 + 4/y-2

THANK YOU!!

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