# 1988 AHSME Problems/Problem 14

## Problem

For any real number a and positive integer k, define

${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$

What is

${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$?

$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$

## Solution

We expand both the numerator and the denominator.

\begin{align*} \binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100} &= \frac{ \dfrac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} }{ \dfrac{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} } \\ &= \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } \end{align*}

Now, note that $-\frac{1}{2}-1=\frac{1}{2}-2$, $-\frac{1}{2}-2=\frac{1}{2}-3$, etc.; in essence, $-\frac{1}{2}-n=\frac{1}{2}-(n+1)$. We can then simplify the numerator and cancel like terms.

\begin{align*} \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } &= \frac{ \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cancel{(\frac{1}{2} - 3)} \cdots (\frac{1}{2} - 100) }{ (\frac{1}{2}) \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cdots \cancel{(\frac{1}{2} - 99)} } \\ &= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\ &= \frac{-\frac{199}{2}}{\frac{1}{2}} \\ &= \boxed{\textbf{(A) } -199.} \end{align*}