1. ## Combinations question

For part (a)

Would i be right to say:

There's sequence of 30 0's and 9 1's.

C(k+n-1,k) = (k+n-1)!/k!(n-1)! =

C(39,30) = 39!/30!9!

The solution someone else gave me is different to my assumptions.

Solution someone else told me:

2. ## Re: Combinations question

Originally Posted by inayat

For part (a)
Would i be right to say:
There's sequence of 30 0's and 9 1's.
C(k+n-1,k) = (k+n-1)!/k!(n-1)! or C(39,30) = 39!/30!9! YES!
b) There are five odd number boxes. If we go ahead and put one ball in to each of those five then distribute the remaining 25 balls into ten boxes in $\dbinom{25+10-1}{25}$ ways. Then odd boxes are nonempty.

c) This can be a trick thought experiment. Zero is an even number, so an empty box contains an even number of balls. Now think if "gluing" the balls into pairs of two each. Now put these 15 identical objects into the ten boxes.
That can be done in $\dbinom{15+10-1}{15}$ ways. Each box contains an number of the original balls.

3. ## Re: Combinations question

Originally Posted by Plato
b) There are five odd number boxes. If we go ahead and put one ball in to each of those five then distribute the remaining 25 balls into ten boxes in $\dbinom{25+10-1}{25}$ ways. Then odd boxes are nonempty.

c) This can be a trick thought experiment. Zero is an even number, so an empty box contains an even number of balls. Now think if "gluing" the balls into pairs of two each. Now put these 15 identical objects into the ten boxes.
That can be done in $\dbinom{15+10-1}{15}$ ways. Each box contains an number of the original balls.
Thanks that helps