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I am having a really hard time to come up with a result for this problem:

In how many ways can you give 3 board games, 2 movie tickets and 4 toys to your 3 friends, assuming that everyone gets at least something? The games, tickets and toys are undistinguishable

I think that the formula I should be using is: C(r + k -1, r) where r = 3 (for 3 kids), but how do I incorporate the difference, if I give ex. 2 games 1 movie ticket to 1 person and then something else to the other 2?

Thank you so much!

T

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1 Answers 1

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I think your best bet is to use inclusion-exclusion to rule out the chance of someone being left out of the distribution. So essentially we start by ignoring the restriction that everyone gets something, and divide out the three types of goods anyway:

$$P_3 = \binom{3+2}2\binom{2+2}2\binom{4+2}2 = \binom 52\binom 42\binom 62 = 10\cdot 6\cdot 15 = 900$$

But then someone might get left out; so we calculate the number of way to divide between two selected people, which is, similarly,

$$P_2 = \binom{3+1}1\binom{2+1}1\binom{4+1}1 = \binom 41\binom 31\binom 51 = 4\cdot3\cdot 5 = 60$$

And when we subtract that off, we will have also oversubtracted the options available when everything goes to one selected person, which is: $$P_1 = \binom 30\binom 20\binom 40 =1$$

of course.

So for the answer desired, note that we can choose two people $3$ ways and one person $3$ ways, so we need the calculation $$R = P_3 - 3P_2 + 3P_1 = 900 - 180 + 3 = 723 \text{ options}$$