# Stars and bars with a twist

Consider $5$ labeled bins and $200$ unlabeled balls. We have the standard stars and bars formula $\binom{200 + 5 - 1}{5 - 1}$ ways to place them in the bins with no restriction.

What if $x_1 \ge x_2 \ge x_3 \ge x_4 \ge x_5 \gt 0$?

• For the now deleted first question, place $2$ balls in the first bin, $3$ balls in the second bin, $4$ balls in the third bin, $5$ balls in the fourth bin, and $6$ balls in the fifth bin. Since no more balls can be placed in the fifth bin and $20$ balls have been placed, the problem reduces to solving the equation $$y_1 + y_2 + y_3 + y_4 = 180$$ in the nonnegative integers. – N. F. Taussig Feb 11 '18 at 22:18
• I realized that question is stupid and have deleted it. – B.Li Feb 11 '18 at 22:21
• Isn't it the coefficient of $z^{300}$ in the Maclaurin expansion of $$\frac1{(1-z)(1-z^2)(1-z^3)(1-z^4)(1-z^5)}\ ?$$ – bof Feb 11 '18 at 23:35
• See integer partitions. Specifically the subheading 'Restricted part size or number of parts'. – N. Shales Feb 12 '18 at 3:45
• answer my own question – B.Li May 15 '18 at 17:29

If $x_1 \ge x_2 \ge x_3 \ge x_4 \ge x_5 \gt 0$, then each $x_i$ is at least $1$. Consider any integer partition of size $5$ on $200$. For any partition, say $$50 + 100 + 25 + 20 + 5 = 200$$ There's always a non-increasing arrangement of the partitions. For example, the above can be arranged to look like $$100 + 50 + 25 + 20 + 5 = 200$$ Hence, $$x_1 = 100, x_2 = 50. x_3 = 25, x_4 = 20, x_5 = 5$$ So the question is an integer partition of size $5$, i.e. $P_5(200)$