PUMaC 2021 · 组合（B 组） · 第 5 题
PUMaC 2021 — Combinatorics (Division B) — Problem 5
题目详情
- Nelson is having his friend drop his unique bouncy ball from a 12 foot building, and Nelson will only catch the ball at the peak of its trajectory between bounces. On any given bounce, 1 there is an 80% chance that the next peak occurs at the height of the previous peak and a 3 20% chance that the next peak occurs at 3 times the height of the previous peak (where the first peak is at 12 feet). If Nelson can only reach 4 feet into the air and will catch the ball as soon as possible, the probability that Nelson catches the ball after exactly 13 bounces is a b c d e 2 × 3 × 5 × 7 × 11 for integers a , b , c , d , and e . Find | a | + | b | + | c | + | d | + | e | .
解析
- Nelson is having his friend drop his unique bouncy ball from a 12 foot building, and Nelson will only catch the ball at the peak of its trajectory between bounces. On any given bounce, 1 there is an 80% chance that the next peak occurs at the height of the previous peak and a 3 20% chance that the next peak occurs at 3 times the height of the previous peak (where the first peak is at 12 feet). If Nelson can only reach 4 feet into the air and will catch the ball as soon as possible, the probability that Nelson catches the ball after exactly 13 bounces is a b c d e 2 × 3 × 5 × 7 × 11 for integers a , b , c , d , and e . Find | a | + | b | + | c | + | d | + | e | . Prooposed by: Rishi Dange Answer: 31 Call an 80% bounce A and a 20% bounce B . It’s easy to see that for 13 bounces to occur before ending, there must be 7 A s and 6 B s. Now we simply need to find the number of orderings of the A s and B s that are possible. The first bounce must be B and the last two must both be A , otherwise Nelson will catch the ball before 13 bounces. For the remaining 10 slots, there must be 5 A s and 5 B s such that within those 10, there is never an instance where the number of A s exceeds the number of B s by 2 or more (not including the bounce before those 10). That way, Nelson won’t catch the ball before the 13 bounce total. This is a simple Catalan counting problem that gives 132 orderings, or C (6). 4 1 7 6 16 1 − 13 0 1 Thus, the final probability is ( ) × ( ) × 132, or 2 × 3 × 5 × 7 × 11 . 5 5