Difference between revisions of "2015 USAMO Problems/Problem 3"
(→Solution) |
(→Solution) |
||
| Line 29: | Line 29: | ||
... (Case II.n+1) Core =S. Then C(S)=1, with all other subsets C(T)=0, there is 1=(nCn)2^0 | ... (Case II.n+1) Core =S. Then C(S)=1, with all other subsets C(T)=0, there is 1=(nCn)2^0 | ||
| − | Combining all the cases, 1+[1+( | + | Combining all the cases, $1+[1+(\dbinom{n}{1})2^{n-1}+(nC2)2^(n-2)+...+(nCn)2^0]=1+3^n is the total number of colorings satisfying the given condition. |
Revision as of 09:20, 3 June 2016
Problem
Let 
, where 
. Each of the 
subsets of 
is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set 
, we then write 
for the number of subsets of T that are blue.
Determine the number of colorings that satisfy the following condition: for any subsets 
and 
of , ![\[f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).\]](http://latex.artofproblemsolving.com/1/e/5/1e5371f0639b2a8ed04d6b74c7401b1d36264652.png)
Solution
Define function: 
if the set T is colored blue, and 
if 
is colored red.
Define the 
.
The empty set is denoted as Nil. ^ denotes intersection; + denotes union. T1<T2 means T1 is a subset of T2 but not =T2. Let Sn={n} are one-element subsets.
Let mCk denote m choose k = m!/(k!(m-k)!)
(Case I) 
)=
. Then for distinct m and k, 
, meaning only if Sm and Sk are both blue is their union blue. Namely 
Similarly, for distinct m,n,k, f(Sm+Sk+Sn)=f(Sm+Sk)f(Sn), C(Sm+Sk+Sn)=C(Sm)C(Sk)C(Sn). This procedure of determination continues to S. Therefore, if T={a1,a2,...ak}, then C(T)=C(Sa1)C(Sa2)...C(Sak). All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition. There are 2^n legit colorings in this case.
(Case II.) f(Nil)=0.
(Case II.1) Core=Nil. Then either (II.1.1) there exist two nonintersecting subsets A and B, C(A)=C(B)=1, but f(A)f(B)=0 which is a contradiction, or (II.1.2) all subsets has C(T)=0, which is easily confirmed to satisfy the condition f(T1)f(T2)=f(T1^T2)f(T1+T2). There is one legitimate coloring in this case.
(Case II.2) Core= a subset of 1 element. WOLG, C(S1)=1. Then f(S1)=1, and subsets containing element 1 may be colored Blue.
f(S1+Sm)f(S1+Sn)=f(S1+Sm+Sn), namely C(S1+Sm+Sn)=C(Sm+S1)C(Sn+S1). Now S1 functions as the Nil in case I, with n-1 elements to combine into a base of n-1 2-element sets, and all the other subsets are determined. There are 2^(n-1) legit colorings for each choice of core. BUt there are nC1 (i.e. n choose 1) =n such cores. Hence altogether there are n2^(n-1) colorings in this case.
(Case II.3) Core = a subset of 2 elements. WLOG, C(S1+S2)=1. Only subsets containing the core may be colored Blue. With the same reasoning as in the preceding case, there are (nC2)2^(n-2) colorings.
... (Case II.n+1) Core =S. Then C(S)=1, with all other subsets C(T)=0, there is 1=(nCn)2^0
Combining all the cases, $1+[1+(\dbinom{n}{1})2^{n-1}+(nC2)2^(n-2)+...+(nCn)2^0]=1+3^n is the total number of colorings satisfying the given condition.
