HMMT 十一月 2010 · 冲刺赛 · 第 9 题

HMMT November 2010 — Guts Round — Problem 9

[ 7 ] How many functions f : { 1 , 2 , . . . , 10 } → { 1 , 2 , . . . , 10 } satisfy the property that f ( i ) + f ( j ) = 11 for all values of i and j such that i + j = 11.

解析

[ 7 ] How many functions f : { 1 , 2 , . . . , 10 } → { 1 , 2 , . . . , 10 } satisfy the property that f ( i ) + f ( j ) = 11 for all values of i and j such that i + j = 11. Answer: 100000 To construct such a function f , we just need to choose a value for f ( x ) from { 1 , 2 , ..., 10 } for each x ∈ { 1 , 2 , ..., 10 } . But the condition that f ( i ) + f ( j ) = 11 whenever i + j = 11 means that f (10) = 11 − f (1) . f (9) = 11 − f (2) . . . . f (6) = 11 − f (5) . This means that once we have chosen f (1), f (2), f (3), f (4), and f (5), the five remaining values of f (6), f (7), f (8), f (9), and f (10) are already determined. The answer is therefore just the number of ways to choose these first five values. Since there are 10 possibilities for each one, we get that the 5 answer is 10 = 100000.