HMMT 十一月 2024 · 冲刺赛 · 第 36 题
HMMT November 2024 — Guts Round — Problem 36
题目详情
- [20] Estimate the value of 40 40 ∑ ∑ 20! · 40! · 40! ( i + j + 18)! · . 100! i ! j !18! i =0 j =0 Submit a positive real number E either in decimal or in a fraction of two positive integers written in 2024 decimal (such as ). If the correct answer is A , your will receive max (20 − ⌊ 5000 · | E − A |⌋ , 0) points. 2025
解析
- [20] Estimate the value of 40 40 X X 20! · 40! · 40! ( i + j + 18)! · . 100! i ! j !18! i =0 j =0 Submit a positive real number E either in decimal or in a fraction of two positive integers written in 2024 decimal (such as ). If the correct answer is A , your will receive max(20 − ⌊ 5000 · | E − A |⌋ , 0) points. 2025 Proposed by: Kevin Zhao Answer: 0.1085859 Solution: Note that 40 40 40 40 40 40 X X X X X X ( i + j + 18)! (98 − i − j )! 98 − i − j = = . i ! j !18! (40 − i )!(40 − j )!18! 40 − i, 40 − j, 18 i =0 j =0 i =0 j =0 i =0 j =0 98 − i − j The multinomial coefficient counts the number of ways to arrange 40 − i red balls, 40 − j 40 − i, 40 − j, 18 blue balls, and 18 green balls in a line. Hence, it also counts the number of ways to arrange 40 red, 40 blue, and 20 green balls such that the first i + 1 balls are i reds followed by a green, and the last j + 1 balls are j blues preceded by a green. Thus, the summation over all possible i and j counts the number of ways to arrange 40 red, 40 blue, and 20 green balls such that all balls before the first green ball are 100! red and all balls after the last green ball are blue. Since there are total permutations of these 20! · 40! · 40! balls, the desired answer is the probability that a random permutation satisfies the above properties. 20 1 Since there are 40 blue and 20 green balls, the probability the first non-red ball is green is = . 60 3 1 Similarly, the probability the the last non-blue ball is green is . If we assume these probabilities are 3 1 independent, we get an answer of , which is accurate enough for 8 points. 9 To get a better estimate, given that the first non-red ball is green, we can find the expected number of red balls that appeared before it. Indeed, this is equivalent to the expected number of red balls before the first non-red ball. There are 60 non-red balls which separate out 61 intervals for the red balls to be in; any given ball is equally likely to be in each interval, so the expected number of red balls in the 40 first interval (i.e., before the first non-red ball) is . 61 40 Thus, once we assume the first non-red ball is green, we expect there to be 40 − red balls left and 61 19 1159 19 green balls. Then, the probability that the last non-blue ball is green is about = . Our 40 3559 59 − 61 1 1159 1159 final estimate is then · = ≈ 0 . 10855, which is accurate enough for the full 20 points. 3 3559 10677