PUMaC 2020 · 组合（A 组） · 第 2 题

PUMaC 2020 — Combinatorics (Division A) — Problem 2

Cary has six distinct coins in a jar. Occasionally, he takes out three of the coins and adds a dot to each of them. Determine the number of orders in which Cary can choose the coins so that, eventually, for each number i ∈ { 0 , 1 , . . . , 5 } , some coin has exactly i dots on it.

解析

Cary has six distinct coins in a jar. Occasionally, he takes out three of the coins and adds a dot to each of them. Determine the number of orders in which Cary can choose the coins so that, eventually, for each number i ∈ { 0 , 1 , . . . , 5 } , some coin has exactly i dots on it. Proposed by Frank Lu Answer: 79200 . Solution: Label the coins 0 , 1 , . . . , 5 by how many dots they end up with; notice that there are 720 ways to make this assignment (depending on how to assign our 6 coins to these dots). Note that, since the sum of the number of dots in total is 15, and we add 3 dots per draw, Cary pulled out coins 5 times. This also means that Cary drew the 5 coin every time. Without loss of generality, assume the 1 coin was drawn in the first pile, and we multiply by 5 later. Now, we have 1 5 , 5 , 5 , 5 , 5 for our draws. Observe that if the 4 coin is never drawn with ( ) 5 the 1 coin, then we have ways to arrange the 2 and 3 coin, all of which work. Otherwise, 2 we have 4 choices for which draw has neither a 1 nor a 4, resulting in an order like the ( ) 3 following (multiplying by 4 later): 145 , 235 , 45 , 45 , 45. Here, we have ways to determine 1 the remaining place in which the 2 coin was drawn. Our total is thus 720 · 5 · (10+4 · 3) = 79200. Note: We initially had 110 as the answer, but this is incorrect since we stated that we had distinct coins. We apologize for the confusion this would have caused.