1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This doesnt take into account those who get two... I guess. $\endgroup$ – user2277550 Sep 22 '16 at 20:43
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.