In how many ways can 30 identical balls be distributed into 6 distinct boxes (numbered box 1, ... , box 6) where each box gets an odd number of balls?

Question

In how many ways can $30$ identical balls be distributed into $6$ distinct boxes (numbered box 1, ... , box 6) where each box gets an odd number of balls?

I have this problem at the moment and the part about there needing to be odd number of balls in every box is really throwing me.

Answer 1

Suppose box $i$ gets $2k_i+1$ boxes. Then you want

$$\sum_{i=1}^6 (2k_i + 1) = 30$$

Or rearranging

$$\sum_{i=1}^6 k_i = \frac{30-6}{2} = 12$$

Now how many solutions are there to $\sum_{i=1}^6 k_i = 12$ ?

Answer 2

My approach is about generating function each box gets odd number of balls

We need to find cofficient of $x^{30}$ in $$(x+x^3+x^5+\dots)^6=\frac{x^6}{(1-x^2)^6}$$

$$=\sum_{k=0}^{\infty}\binom{k+5}{k}x^{2k+6}$$

Cofficient of $x^{30}$ is $$\binom{17}{12}$$