Exercise 1.59 (Distributing chocolate bars)

Question

There are 15 chocolate bars and 10 children. In how many ways can the chocolate bars be distributed to the children, in each of the following scenarios?
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(a) The chocolate bars are fungible (interchangeable).
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(b) The chocolate bars are fungible, and each child must receive at least one.
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(c) The chocolate bars are not fungible (it matters which particular bar goes where).
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(d) The chocolate bars are not fungible, and each child must receive at least one.

fifthist x · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item $\binom{15+9}{9}$

\item $\binom{5+9}{9}$

\item 
Each of 15 bars can be given to any of 10 children, so by ordered sampling with replacement formula, we have $10^{15}$ combinations.

\item
To count amount of suitable combinations, we can subtract amount of combination, where at
least one child doesn't get any bars (is example of inclusion-exclusion usage case) from
total amount of combinations.
$$10^{15} - \sum_{i=1}^{9}\left(-1\right)^{i+1}\binom{10}{i}\left(10-i\right)^{15}.$$
\end{enumerate}

Exercise 1.59 (Distributing chocolate bars)

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Exercise 1.59 (Distributing chocolate bars)

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