Combinatorics Question

May 2017
6
0
United States
I'm having issue solving this problem




If they were said that xi are non-negative integers I would know how to solve it. It would be n = 32, k = 6, and then (32+6-1) chooses (6-1). But I'm not sure how to apply any of these constraints. My professor didn't go into detail about it and only gave us one example in class. I would really appreciate it if you could provide me with some explanation too! Thank you!
 

Debsta

MHF Helper
Oct 2009
1,346
623
Brisbane
I'm having issue solving this problem




If they were said that xi are non-negative integers I would know how to solve it. It would be n = 32, k = 6, and then (32+6-1) chooses (6-1). But I'm not sure how to apply any of these constraints. My professor didn't go into detail about it and only gave us one example in class. I would really appreciate it if you could provide me with some explanation too! Thank you!
Can't see the problem!
 

Plato

MHF Helper
Aug 2006
22,492
8,653
I'm having issue solving this problem

If they were said that xi are non-negative integers I would know how to solve it. It would be n = 32, k = 6, and then (32+6-1) chooses (6-1). But I'm not sure how to apply any of these constraints. My professor didn't go into detail about it and only gave us one example in class. I would really appreciate it if you could provide me with some explanation too! Thank you!
You are correct about the total: $\dbinom{32+6-1}{32}$
a) $\dbinom{16+6-1}{16}$ Now why sixteen? Zero is even. If we think about gluing the balls(the ones) together in pairs there are sixteen pairs. Putting those into the six boxes we have even solutions.

b) Subtracting the answer in a) from the total we get the odd solutions.

c) $\dbinom{26+6-1}{26}$ Well $32-6=26$ we go ahead and put a one into each cell so each cell has at least one (each solution is positive, not zero).

d) $\dbinom{32+6-1}{32}-\dbinom{29+6-1}{29}$ I will let you explain that answer. Please post it.