\problemdata       % Takes 5 *mandatory* arguments
  {candy-counting}         % latex-friendly label for the prob.
  {combinatorics}          % The topic(s) of the problem content.
  {}                       % Source (if known)
  {S02 q3-1}               % Usage (list ones you are aware of).
  {Nick Matsakis, S02}     % Last revision info (author, date).
\begin{problem}[20]
Four (distinguishable) children approach you on Halloween, requesting candy.
You have 10 pieces of candy to distribute among them, and you don't
want to keep any candy for yourself. Determine how many different ways you
can distribute the 10 pieces under the following conditions. \emph{You may
specify your answer in terms of binomial coefficients.}

\begin{problemparts}

%%%%%%%%
\problempart[6] 
All the pieces are the same (Reese's Peanut Butter Cups\texttrademark).

\solution[\vspace*{3.5cm}]{ This is an indistinguishable balls into
distinguishable urns problem. The candy pieces are the balls and the
children are the urns. Or, in the "stars and bars" forumulation, the
10 stars are the candy pieces and the $4-1$ bars are the separators
between the children.

$$
{10+4-1 \choose 10} = {13 \choose 10} = \frac{13!}{3! \cdot 10!} = 286
$$
}

%%%%%%%%
\problempart[6] 
All 10 pieces of candy are different. 
\solution[\vspace*{3.5cm}]{
You can think of this as putting 10 distinct balls (the candy) into four
distinct urns (the children). There are 4 choices for the first ball, 4
for the second, and so forth.
$$
4^{10}
$$
}

%%%%%%%%
\problempart[8] 5 pieces are Reese's \texttrademark and the other
5 are Nestl\'{e}'s Crunch\texttrademark.

\solution[\vspace*{5cm}]{ We count the number ways to distribute the
Peanut-Butter cups to children, multiplied by the number of ways to
distribute the Crunch bars.
$$
{5+4-1 \choose 5}\cdot {5+4-1 \choose 5} = {8 \choose 5}^2 =
\left(\frac{8!}{3!5!}\right)^2 = 3136
$$
}

\iffalse
%%%%%%%%
\problempart[6] 
All the pieces are the same, but no child gets more than 4 pieces.

\solution[\vspace*{2cm}]{
We use the formula for 10-combinations of 4 elements with
repetition with each element occuring fewer than 5 times, and get:

$$
{13 \choose 3} - {4 \choose 1}{8 \choose 3} + {4 \choose 2}{3
  \choose 3} = 68
$$
}
\fi
\end{problemparts}
\end{problem}
