1988 AMC 12 Problem 14

Question

For any real number

a

and positive integer

k,

define

\binom ak= \frac{\begin{gathered} a(a-1)(a-2)\cdots\\ {}\cdot(a-(k-1)) \end{gathered}} {\begin{gathered} k(k-1)(k-2)\cdots\\ {}\cdot(2)(1) \end{gathered}}.

What is

\binom{-\frac12}{100}\div\binom{\frac12}{100}\ ?

-199

-197

-1

197

199

Cancel the identical denominator

100!

Write the remaining factors as consecutive odd integers and look for telescoping cancellation

After canceling

100!,

the numerator factors have magnitudes

1,3,\ldots,199,

while the denominator factors have magnitudes

1,1,3,\ldots,197.

All common odd factors cancel, leaving magnitude

199.

The numerator has

100

negative factors and the denominator has

99,

so the quotient is

-199.

Thus the correct answer is A.

Problem 14 in other years