HMMT 二月 2016 · COMB 赛 · 第 8 题
HMMT February 2016 — COMB Round — Problem 8
题目详情
英文原题
- Let X be the collection of all functions f : { 0 , 1 , . . . , 2016 } → { 0 , 1 , . . . , 2016 } . Compute the numberof functions f ∈ X such that
( )
( ) ( )
max min max( f ( i ) , g ( i )) − max min( f ( i ) , g ( i )) = 2015 .
g ∈ X 0 ≤ i ≤ 2016 0 ≤ i ≤ 2016
解析
英文解析
- Let X be the collection of all functions f : { 0 , 1 , . . . , 2016 } → { 0 , 1 , . . . , 2016 } . Compute the numberof functions f ∈ X such that
( )
( ) ( )
max min max( f ( i ) , g ( i )) − max min( f ( i ) , g ( i )) = 2015 .
g ∈ X 0 ≤ i ≤ 2016 0 ≤ i ≤ 2016
Proposed by:
2017 2017
Answer: 2 · (3 − 2 )
For each f, g ∈ X , we define
( ) ( )
d ( f, g ) := min max( f ( i ) , g ( i )) − max min( f ( i ) , g ( i ))
0 ≤ i ≤ 2016 0 ≤ i ≤ 2016
Thus we desire max d ( f, g ) = 2015.
g ∈ X
First, we count the number of functions f ∈ X such that
∃ g : min max { f ( i ) , g ( i ) } ≥ 2015 and ∃ g : min max { f ( i ) , g ( i ) } = 0 .
i i
That means for every value of i , either f ( i ) = 0 (then we pick g ( i ) = 2015) or f ( i ) ≥ 2015 (then we
2017
pick g ( i ) = 0). So there are A = 3 functions in this case.
Similarly, the number of functions such that
∃ g : min max { f ( i ) , g ( i ) } = 2016 and ∃ g : min max { f ( i ) , g ( i ) } ≤ 1
i i
2017
is also B = 3 .
Finally, the number of functions such that
∃ g : min max { f ( i ) , g ( i ) } = 2016 and ∃ g : min max { f ( i ) , g ( i ) } = 0
i i
2017
is C = 2 .
Now A + B − C counts the number of functions with max d ( f, g ) ≥ 2015 and C counts the numberg ∈ X
2017 2017
of functions with max d ( f, g ) ≥ 2016, so the answer is A + B − 2 C = 2 · (3 − 2 ).
g ∈ X