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If I have 1000 identical objects that I need to distribute over 5 bins with the restraint that there can't be less than 40 in each bin, in how many ways can the distribution be done in?

I think that I first need to take $$\frac{\binom{1000}{120}}{5}$$

Now we have distributed 40 objects into each container. The rest comes from doing a partition, with the remaining objects.

$$\displaystyle\binom{n+k-1}{k-1} = $$ $$\displaystyle= \frac{(n+k-1)!}{(k-1)!(n-k)!}$$

And finally I combine the two calculations by multiplying them?

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  • $\begingroup$ If you put $40$ objects in each of the five containers initially, that leaves you with $800$ identical objects to distribute. You need to solve the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 800$$ in the nonnegative integers. $\endgroup$ – N. F. Taussig Feb 5 '17 at 17:58
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They are identical objects, and (presumably) distinct bins.

All you need to do is to first put $40$ in each bin, and then apply stars and bars for the remaining $880$ to get $\binom{880+5-1}{5-1}$ ways

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