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2000 AIME II Problems/Problem 3

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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 a two cards from the 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}.

See also

2000 AIME II (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Preparing for MATHCOUNTS or the AMC contests, and having a tough time with number theory problems? Read Art of Problem Solving's Introduction to Number Theory by Mathew Crawford.
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