HMMT 二月 2013 · 冲刺赛 · 第 29 题
HMMT February 2013 — Guts Round — Problem 29
题目详情
英文原题
- [ 20 ] Let A , A , ..., A be finite sets of size 2012 and let B , B , ..., B be finite sets of size 2013 such
1 2 m 1 2 mthat A ∩ B = ∅ if and only if i = j . Find the maximum value of m .
i j
解析
英文解析
- [ 20 ] Let A , A , ..., A be finite sets of size 2012 and let B , B , ..., B be finite sets of size 2013 such
1 2 m 1 2 mthat A ∩ B = ∅ if and only if i = j . Find the maximum value of m .
i j
( )
4025
Answer: In general, we will show that if each of the sets A contain a elements and if eachi
2012
( )
a + bof the sets B contain b elements, then the maximum value for m is .
Let U denote the union of all the sets A and B and let | U | = n . Consider the n ! orderings of theaji jelements of U . Note that for any specific ordering, there is at most one value of i such that all theelements in A come before all the elements in B in this ordering; this follows since A shares at leasti i jone element with B and B shares at least one element with A for any other j 6 = i .
i j i
On the other hand, the number of ways to permute the ( a + b ) elements in A ∪ B so that all thei ielements in A come first is equal to a ! b !. Therefore, the number of permutations of U where all theielements in A come before all the elements in B is equal to:
i ia ! b ! n !
n ! · = ( )
a + b
( a + b )!
Summing over all m values of i , the total number of orderings where, for some i , the elements in Aaicome before B is equal toi
Guts Roundn ! m
( )
a + ba
( )
a + b
But there are at most u ! such orderings, since there are u ! total orderings, so it follows that m ≤ .
Equality is attained by taking U to be a set containing ( a + b ) elements, letting A range over allaia -element subsets of U , and letting B = U \ A for each i .
i i