INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

December 3, 2012

RMO 2012 solution to Question No. 4

4. Let X = {1, 2, 3, ... , 10}. Find the number of pairs {A, B} such that A ⊆ X, B ⊆ X, A ≠ B and A∩B = {5, 7, 8}.
 
Solution:
 
First we put 5, 7, 8 in each of A and B.
 
We are left out with 7 elements of X.
 
For each of these 7 elements there are three choices:
a) it goes to A
b) it goes to B
c) it goes to neither A nor B
 
Hence there are total (3^7) = 2187 choices. From these 2187 cases we delete that one case where all of the seven elements goes to neither A nor B as A≠ B thus giving 2187 -1 = 2186 cases.
 
Since A and B is unordered (that is A= {5, 7, 8, 1, 2} , B = {5, 7, 8, 4} is the same as B= {5, 7, 8, 1, 2} , A = {5, 7, 8, 4} ) we take half of these 2186 cases that is 1093 cases.
 
Hence there are 1093 such pairs.

2 comments on “RMO 2012 solution to Question No. 4”

  1. Dear Soumik,The official solution and our solution (and hence the result) differs because they have taken A and be as ordered, whereas we took it as unordered. Otherwise both solutions are same.

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
enter