AboutSteve Holleran Expertise I can help with all math questions from basic math to Calculus.
Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.
Experience 33 years teaching experience in NJ public schools
Education/Credentials B.S. Mathematics : Wake Forest University 1972
M.S. Mathematics : Monmouth University 1981
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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.
