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HMMT 十一月 2010 · 冲刺赛 · 第 9 题

HMMT November 2010 — Guts Round — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 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

英文解析

  1. [ 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 thatf (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 off (6), f (7), f (8), f (9), and f (10) are already determined. The answer is therefore just the numberof ways to choose these first five values. Since there are 10 possibilities for each one, we get that theanswer is 10 = 100000.5