Expert: Steve Holleran - 5/5/2008

Question
Hello! The expert can't answer your question.

Your Question was:

We have 7 white balls and 6 red balls. In how many ways

can we put them in 4 different boxes?

Because Hi Magnus,



This is not exactly my specialty, but I have some
additional questions about the whole set up of the
problem.



Do you have to use all 13 balls? And does each box have to
have at least 1 ball in it, or can you leave a box, or
boxes, empty?



In any case, this is a pretty formidable problem, and I'm
not sure I'll be of much help, but if you can clarify the
situation a little, I'll try.



Let me know what you think,

Steve

Expert: Steve Holleran

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Hey Steve

You have to use all the balls, and boxes can be empty.
Im not sure, but i think i have figure out the problem
now.

If we first figure out in how many ways we can choose the
balls, than it would be 13!/(7!*6!). If we multiple that
whit how many different ways we can put 13 equal balls in
4 boxes, that would be 4^13.

So i think the answer is (13!/(7!*6!))*4^13. but not quit
sure.

thanks for help.

Answer
Magnus,

Thanks for your follow-up.

I think, however, there are some inaccuracies here.

The expression you give, 13! / (7! * 6!) is the number of DISTINGUISHABLE ARRANGEMENTS of the 13 balls, not the number of ways you  can choose the balls.

And I think the number of ways to place 13 balls in 4 boxes would be
13^4 , not 4^13.

Also, we haven't accounted for  the fact that any box or boxes can be empty, and that there's no restriction on how many balls can be in each box.  

Just consider, for example, you can leave boxes A, B, and C empty and then put all 13 balls in box D.  Or,you can leave A,B, and D empty and put all 13 in box C, etc.  Or, you can leave A and B empty, and divide the balls into boxes C and D; and on and on.

I honestly don't know how to approach this problem.  I'll play with it for another while, but I don't think I'll get too far.  Sorry to disappoint.

Steve