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Is this similar if not the same to the distribution of pennies since they did not mentioned that the presents are different? In which case we can say that there is $ \binom{n-1}{k-1}$ choices or is my thinking wrong? (where $n$ are the presents and $k$ are the children)

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  • $\begingroup$ If the presents are indistinguishable, then that's the right answer. If they are distinguishable, then clearly there are more ways. I don't know a better way to calculate that than to calculate the number of ways to distribute presents without limitations, then subtract the number of ways to distribute where at least one kid gets nothing. That's a messy inclusion-exclusion thing. $\endgroup$ – Arthur Mar 22 '17 at 7:00
  • $\begingroup$ For the sake of covering every outcome what would it look like if the presents are distinguishable would it be $15 \times 14 \times 13 \times 12 \times 11$ $\endgroup$ – Squanchinator Mar 22 '17 at 7:07
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    $\begingroup$ If the presents are distinguishable, then if children can get 0 presents (sad!), the answer is $5^{15}$, much larger than $15\times14\times13\times12\times11$. I think the inclusion–exclusion, under the restriction that each chlid gets at least 1 present, works out to$$5^{15} - \binom514^{15} + \binom523^{15} - \binom532^{15} + \binom541^{15}.$$ $\endgroup$ – Greg Martin Mar 22 '17 at 7:40
  • $\begingroup$ That's freaking awesome. These counting problems can get pretty weird. Would you mind explaining your thought process for the above if you have the time. $\endgroup$ – Squanchinator Mar 22 '17 at 8:23
  • $\begingroup$ It's what you get if you count the different categories "Exactly $k$ kids get no presents" and "At most $k$ kids get no presents", starting with $k=4$ and working your way down to $k=0$. $\endgroup$ – Arthur Mar 22 '17 at 8:32

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