Imagine that you have a warehouse full of 6 different types presents. You want to send 1 or 2 presents to each of 10 different coworkers. How many different ways are there to do this?

Remark: If one of the coworkers gets two presents, they must be different.

Is it correct to think of this situation as distributing 10 identical presents to 6 different types?

This would imply: $${n + k-1 \choose k-1} = {10 + 6-1 \choose 6-1} = {15 \choose 5}$$

I feel that this is incorrect because I don't think that this assures that each individual gets a minimum of 1 present. It would be great if someone could explain the logic of this problem to me. Thank you.

  • $\begingroup$ This doesnt take into account those who get two... I guess. $\endgroup$ – user2277550 Sep 22 '16 at 20:43
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    $\begingroup$ Perhaps I am not understanding the question. Just consider the case where every employee gets $2$. There are $\frac {6\times 5}2=15$ ways to choose a pair of presents and I get one such choice for each employee...thus this case contributes $15^{10}=576650390625$. Whereas $\binom {15}5 = 3003$. Or have I misunderstood? $\endgroup$ – lulu Sep 22 '16 at 20:47
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    $\begingroup$ Even the case wherein each employee gets exactly $1$ present contributes $6^{10}=60466176$. $\endgroup$ – lulu Sep 22 '16 at 20:50
  • $\begingroup$ @lulu I think that you are correct. Your stating that there is $6 \choose 2$ ways to pick $2$ presents from $6$ for the first person, and the same for all the rest which does imply $15^{10}$ because these choices are successive with replacement? But now what I don't understand is what if now they can recieve one OR two presents? $\endgroup$ – RedShift Sep 22 '16 at 20:58
  • $\begingroup$ The posted solution from @user2277550 handles it correctly. $\endgroup$ – lulu Sep 22 '16 at 21:47

2 different presents can be chosen in $ 6 \choose 2$ ways. 1 can be chosen in $6\choose1$ ways.

So the final answer would be $$ \left({6 \choose 2} + {6 \choose 1}\right)^{10} $$

This takes care of all the cases I guess.

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    $\begingroup$ Note: I took the liberty of making your parentheses larger, please undo what I did if you prefer it otherwise. (+1) for the full solution. $\endgroup$ – lulu Sep 22 '16 at 22:04

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