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We have a collection of 40 boxes, each is one of four sizes (evenly distributed 10 of each): sizeA, sizeB, sizeC, sizeD.

Our person is presented with 1 box of each size, they must select one box from the different sizes, then the remaining three will be taken away and four new boxes of each size is presented.

I know this gives me $4^{10}$ possibilities.

But consider this now. Of the 10 boxes the person must end with these sizes:

1 sizeA, 
2 sizeB, 
3 sizeC, 
1 sizeD, 
3 of any size.

I'm trying to figure out based on this how many possibilities I would have now? I'm just a web developer trying a side project thing and I'm very fresh with this.

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1 Answer 1

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Imagine that this person is planning out his choices in advance. He has 10 slots and has to assign each a letter- A, B, C, or D.

First, he must put an A somewhere. He has 10 choices.

Second, he must put two Bs somewhere. He has $9\choose2$ (9 choose 2) choices, no matter what choice he made in the first step. (Take a look here if you're unfamiliar with "n choose k", otherwise known as the binomial coefficient).

Third, he must put 3 Cs somewhere. He has $7\choose3$ choices.

Fourth, he must put a D somewhere. He has 4 choices.

Finally, he has 3 remaining blanks. He can fill each of these with any letter he likes. He thus has $4^3$ choices.

This means the number of choices he has is:

$10\times{9\choose2}\times{7\choose3}\times4\times4^3=3225600$

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  • $\begingroup$ +1 for a nice clear exposition. Of course, this answer is only correct if the placement/order of the boxes matters. If not, all that matters is the final three unknown choices, for which there are only ${4\choose3}=4$ possibilities. But I think your interpretation of the question is probably correct. $\endgroup$ Commented Feb 14, 2012 at 9:53
  • $\begingroup$ Agreed- I interpreted the order mattering because the question stated that the "choose any box" version had $4^{10}$ options $\endgroup$ Commented Feb 14, 2012 at 10:12
  • $\begingroup$ @DavidRobinson That was exactly what I was looking for thanks David. $\endgroup$ Commented Feb 14, 2012 at 21:52

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