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In how many ways can I distribute $100$ identical balls into $6$ different boxes so that no box is left empty and every box contains even number of elements?

Firstly, I placed $2$ balls in every box, which covers the condition that no box is left empty and since they all have to contain even number of elements I chose $2$ balls. Now there are $88$ balls left which I have to distribute. I just don't know how to cover the condition that every box should have even number of elements? Can I for example, look at two balls as "one" ball, and distribute the "$44$" balls into $6$ boxes? Then I could use Stirling number of the second kind...

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    $\begingroup$ Why not count the ways to distribute $50$ Big Balls into $6$ boxes (where a Big Ball means a pair of regular balls). $\endgroup$
    – lulu
    Nov 4 '20 at 16:58
  • $\begingroup$ Exactly, thanks a lot! $\endgroup$
    – untitled
    Nov 4 '20 at 16:59
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    $\begingroup$ Your method of getting even is correct. But the Stirling number of the second kind is applicable when you are making indistinguishable groups of distinct items. In this case, just use star and bar method. $\endgroup$
    – Math Lover
    Nov 4 '20 at 17:16
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Are you familiar with stars and bars method?

So we have to find the number of 6-tuples $(x_1,...x_6)$ such that each $x_i$ is even number grater than $0$ and their sum is 100. So we can write each $x_i=2y_i$ and we have $$y_1+y_2+...+y_6 = 50$$

If we write $z_i= y_i-1$ we have $$z_1+z_2+...+z_6 = 44$$ and that we can write on $${44+5\choose 5}$$ ways.

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  • $\begingroup$ Just a quick question, shouldn't it be $z_i=y_i-2$ since $y_i \geq 2$? $\endgroup$
    – untitled
    Nov 11 '20 at 17:44
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    $\begingroup$ No, $x_i\geq 2$, so $y_i\geq 1$. $\endgroup$
    – Aqua
    Nov 11 '20 at 17:54

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