HMMT 二月 2013 · COMB 赛 · 第 4 题
HMMT February 2013 — COMB Round — Problem 4
题目详情
英文原题
How many orderings ( a1, . . . , a 8) of (1 ,2, . . . , 8) exist such that a1−a2+a3−a4+a5−a6+a7−a8= 0?
解析
英文解析
- How many orderings ( a , . . . , a ) of (1 , 2 , . . . , 8) exist such that a − a + a − a + a − a + a − a = 0?
1 8 1 2 3 4 5 6 7 8
Answer: 4608 We can divide the numbers up based on whether they have a + or - beforethem. Both the numbers following +’s and -’s must add up to 18. Without loss of generality, we canassume the +’s contain the number 1 (and add a factor of 2 at the end to account for this). Thepossible 4-element sets containing a 1 which add to 18 are { 1 , 2 , 7 , 8 } , { 1 , 3 , 6 , 8 } , { 1 , 4 , 5 , 8 } , { 1 , 4 , 6 , 7 } .
Additionally, there are 4! ways to order the numbers following a + and 4! ways to order the numbersfollowing a − . Thus the total number of possibilities is 4 × 2 × 4! × 4! = 4608.