2000 AIME II Problems/Problem 3

Problem

A deck of forty cards consists of four $1$ 's, four $2$ 's,..., and four $10$ 's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

There are ${38 \choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\frac{54+1}{703} = \frac{55}{703}$ , and $m+n = \boxed{758}$ .

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2000 AIME II Problems/Problem 3

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