2018 AIME I Problem 3

Below is the professionally curated solution for Problem 3 of the 2018 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AIME I solutions, or check the answer key.

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Concepts:basic probabilityarrangements with restrictionscasework

Difficulty rating: 2400

3.

Kathy has 55 red cards and 55 green cards. She shuffles the 1010 cards and lays out 55 of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

There are 109876=3024010 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240 equally likely ordered layouts of 55 of the 1010 distinct cards. Kathy is happy exactly when the color pattern consists of one block of reds and one block of greens: the patterns are RRRRR, GGGGG, and the eight mixed patterns RrG5r\text{R}^r\text{G}^{5-r} and G5rRr\text{G}^{5-r}\text{R}^r for r=1,2,3,4.r = 1, 2, 3, 4.

A pattern using rr red and 5r5 - r green positions can be filled in 5!(5r)!5!r!\frac{5!}{(5-r)!} \cdot \frac{5!}{r!} ways (ordered choices of which red cards and which green cards appear). For r=5,4,3,2,1,0r = 5, 4, 3, 2, 1, 0 these counts are 120,120, 600,600, 1200,1200, 1200,1200, 600,600, 120.120. The happy layouts number 120+120+2(600+1200+1200+600)=7440.120 + 120 + 2\,(600 + 1200 + 1200 + 600) = 7440.

The probability is 744030240=31126,\frac{7440}{30240} = \frac{31}{126}, so m+n=31+126=157.m + n = 31 + 126 = 157.

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