\problemdata       % Takes 5 *mandatory* arguments
  {false-counting}         % latex-friendly label for the prob.
  {combinatorics, counting}% The topic(s) of the problem content.
  {}                       % Source (if known)
  {S02 q3-3}               % Usage (list ones you are aware of).
  {Nick Matsakis, S02}     % Last revision info (author, date).
\begin{problem}[15]
The Research division of an Electronics Company has 22 regular
employees: 14 MIT graduates (who do the actual research) and 8 Harvard
graduates (who mostly help write the papers at the end). 

To his dismay, Dilbert, the Head of the team, was told by the
company Boss that he must dispatch 6 of these 22 employees to the 
Development division.

To be fair, Dilbert decides he should include \emph{at least} one MIT
graduate in that team of 6. This leads to a lot of possible teams, and
Dilbert decides he'd better count how many.

\begin{quote}
%lets the inline combinations breath
\renewcommand{\baselinestretch}{1.22}\normalsize
(i) Since the team must have at least one MIT graduate, there are
14 ways to select this MIT graduate. This leaves $13+8=21$ people from
which to select the remaining 5 team members, and there are ${21
\choose 5}$ ways to choose those 5 people. Hence there are $14
\cdot {21 \choose 5} = 284886$ teams with at least one MIT graduate.
\end{quote}

To verify his answer, Dilbert decides to count again in a different way.

\begin{quote}
\renewcommand{\baselinestretch}{1.22}\normalsize 
%lets the inline combinations breath 
(ii) There are ${22 \choose 6}$ different possible teams, and ${8
\choose 6}$ teams of just Harvard graduates. So there are ${22
\choose 6} - {8 \choose 6} = 74585$ different possibilities for
teams with at least one MIT graduate.
\end{quote}

Now Dilbert is worried...

\begin{problemparts}

\problempart[5] \label{incorrect}Which count is \emph{incorrect}? (circle one)

\hspace{.5in} (i) only
\hspace{.5in} (ii) only
\hspace{.5in} both  

\solution{(i) only}

\problempart[10] Carefully explain what is wrong with the incorrect
count(s) you indicated in part (\ref{incorrect})

\solution{The first count overcounts the number of teams with
MIT graduates. Consider a team with MIT graduates, Alyssa and Ben on it. This
team is counted twice, once when Alyssa is selected as the first
MIT graduate and a second time when Ben is selected as the first MIT graduate.
Similarly, teams with three (or more) MIT graduates are overcounted by three
(or more) times.  This cannot be fixed simply with the division rule for this
reason. }

\end{problemparts}
\end{problem}
