Combination problem with distributing

Question

I've been trying to do last exercise, but I can't figure out method to solve it. I read a book and searched for it in the internet, but I couldn't find exactly what I am looking for. Could you guys help me please? Question : We have 5 distinguishable ("different" if I translated this word correctly) mascots and we want to distribute them to 3 children so that each of the children get AT LEAST one mascot. I would really appriciate your help!

Answer 1

We give two solutions, one using a division into cases, and the other using Inclusion/Exclusion.

A Cases solution: Either (i) one lucky kid will get three mascots, and the other two one each or (ii) one unlucky kid will get one mascot, and the other two will get two each.

Case (i): The kid who gets $3$ can be chosen in $\binom{3}{1}$ ways. For each way, the mascots she gets can be chosen in $\binom{5}{3}$ ways. Now the mascot for the oldest remaining kid can be chosen in $\binom{2}{1}$ ways, and it's over. Number of Case (i) ways: $\binom{3}{1}\binom{5}{3}\binom{2}{1}$.

Case (ii): The kid who gets $1$ can be chosen in $\binom{3}{1}$ ways. For each way, her mascot can be chosen in $\binom{5}{1}$ ways. Now that this is done, the $2$ mascots the oldest remaining kid gets can be chosen in $\binom{4}{2}$ ways. Number of Case (ii) ways: $\binom{3}{1}\binom{5}{1}\binom{4}{2}$.

Finally, add.

Inclusion/Exclusion: We describe fairly briefly an Inclusion/Exclusion approach that is more interesting, and works reasonably well even with substantially larger numbers.

If we do not insist that each kid get at least one mascot, there are $3^5$ ways to distribute the mascots between the kids, since for each mascot there are $3$ choices for who it goes to.

Now we must remove the bad distributions, where one or more of the kids gets shut out.

There are $2^5$ ways kid 1 gets shut out, also $2^5$ for kid 2, and $2^5$ for kid 3. However, if we add these numbers, we double-count, for example, the distribution in which kids 1 and 2 get shut out, and so on. So the number of bad distributions is $3\cdot 2^5-3$. It follows that the number of good distributions is $$3^5-3\cdot 2^5+3.$$

Answer 2

You have two possible distributions:

• One child gets three mascots; the other two get one.
• Two children get two mascots; the third gets one.

For the first case, pick the child that gets three mascots, then pick the three mascots that child gets: $3 \times {5 \choose 3} = 30$. Since there are two choices for dividing up the remaining two mascots to the remaining two children in each case, the total number of possibilities is $30 \times 2 = 60$.

I'll give you a shot at the second case. Good luck!