2000 AIME II Problem 3

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3.

A deck of forty cards consists of four 11's, four 22's, ,\ldots, and four 1010'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/nm/n be the probability that two randomly selected cards also form a pair, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 758
Concepts:basic probabilitycombinations
Difficulty rating: 2020
Solution:

After the matching pair is removed, 3838 cards remain: nine numbers with four cards each and one number with only two cards. The number of ways to draw a pair is 9(42)+(22)=54+1=55,9\binom{4}{2} + \binom{2}{2} = 54 + 1 = 55, out of (382)=703\binom{38}{2} = 703 equally likely draws.

Since 703=1937703 = 19 \cdot 37 shares no factor with 55=511,55 = 5 \cdot 11, the probability 55703\frac{55}{703} is in lowest terms, and m+n=55+703=758.m + n = 55 + 703 = 758.

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