# Number of possible distributions of balls, under constraints

In how many different ways can 17 identical red balls and 10 identical white balls be distributed into 4 distinct boxes such that the number of red balls is greater than white balls in each box?

I am not getting any clues how to begin. I tried to apply the same procedure as one would use without the constraints and hoped to find some pattern, without any success.

Strategy:

1. Distribute the white balls without restriction. If $$w_i$$ represents the number of white balls placed in the $$i$$th box, then $$w_1 + w_2 + w_3 + w_4 = 10 \tag{1}$$ is an equation in the nonnegative integers.
2. For each such distribution of white balls, begin by placing $$w_1 + 1$$ red balls in the first box, $$w_2 + 1$$ red balls in the second box, $$w_3 + 1$$ red balls in the third box, and $$w_4 + 1$$ red balls in the fourth box to satisfy the requirement that the number of red balls is greater than the number of white balls in each box. Once you have done so, you will have distributed $$14$$ of the $$17$$ red balls. Distribute the remaining three red balls to the four boxes without restriction. The number of ways to do this is the number of solutions of the equation $$r_1 + r_2 + r_3 + r_4 = 3 \tag{2}$$ in the nonnegative integers.

The number of solutions of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = k$$ in the nonnegative integers is the number of ways $$n - 1$$ addition signs can be inserted into a row of $$k$$ ones, which is $$\binom{k + n - 1}{n - 1}$$ since we must choose which $$n - 1$$ of the $$k + n - 1$$ positions required for $$k$$ ones and $$n - 1$$ addition signs will be filled with addition signs.

hint:

distribute the 10 white balls, then distribute the "excess" of 7 balls.

• Actually excess of $3$ balls. Placement of $14$ red balls is already fixed based on distribution of $10$ white balls. Sep 8 '20 at 10:58
• well, excess of 7 balls with at least 1 ball per bin because of the "greater" and not "greater or equal" Sep 8 '20 at 14:59