HMMT 二月 2006 · GEN2 赛 · 第 8 题
HMMT February 2006 — GEN2 Round — Problem 8
题目详情
英文原题
- Six people, all of different weights, are trying to build a human pyramid: that is, they get into theformation
B CA
D E F
We say that someone not in the bottom row is “supported by” each of the two closest people beneathher or him. How many different pyramids are possible, if nobody can be supported by anybody oflower weight?
解析
英文解析
- Six people, all of different weights, are trying to build a human pyramid: that is, theyget into the formation
B CA
D E F
We say that someone not in the bottom row is “supported by” each of the two closestpeople beneath her or him. How many different pyramids are possible, if nobody can be supported by anybody of lower weight?
Answer: 162
Solution: Without loss of generality, let the weights of the people be 1, 2, 3, 4, 5,
and 6. Clearly we must have A = 1. Then, equally clearly, either B or C must be 2.
Suppose B = 2: Then either C or D must be 3. If C = 3, we have 3! = 6 possibilitiesto fill the bottom row. If D = 3, then C = 4 and we have 2! = 2 possibilities to fill Eand F . Altogether there are 6 + 2 = 8 possibilities in this case.
Suppose C = 2: then, similarly, there are 8 possibilities here.
Altogether there are 8 + 8 = 16 possibilities.