\problemdata                                                             
        {counting-22} % latex-friendly label for the prob.
        {Basic counting; product rule; sum rule; false arguments} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S02 cp7M; F01} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           

In each case, state whether the argument is correct or incorrect. If
incorrect, try to fix it. If correct, prove it.
\begin{problemparts}

\problempart Since there are 26 letters in the English alphabet, the
Product Rule implies that the number of four-letter English words is
$26^4$.

\solution{Incorrect. Not every four letter sequence is an English
word, \eg ``xxxx'' is not English. The best we can do here is say that
the number of four-letter English words is less than $26^4$.}

\problempart The number of 4-digit numbers which have a digit 7
somewhere is $4 * 10^3$, since by the sum rule, we must add the counts
for each position the 7 could be in; and since for each such position,
there are $10^3$ possible numbers by the product rule. (Assume that
the numbers can have leading zeros and that each digit can take values
0-9.)

\solution{Incorrect, since it double counts numbers containing more
than one 7 (for example 7707 is counted 3 times). The sum rule does
not apply if the sets being added are not disjoint. There are many
different ways of counting this, however one very elegant way of
counting this is by counting the complement --- the number of 4-digit
numbers with at least one 7 is just the number of all possible 4-digit
numbers minus the number of those which do \emph{not} contain 7,
namely, $10^4 - 9^4$. This is a common way of counting things where
{\em at least one} special event occurs.}

\end{problemparts}
\end{problem} 
                                 