Five distinct n-digit binary numbers are such that for any two numbers chosen from them the digits will coincide in precisely m places. There is no place with the common digit for all the five numbers. At least one of the binary numbers contains leading zero.

Prove that 2/5 ≤ m/n ≤ 3/5.

For any n there are 2^n binary numbers.  So there are C(2^n,5) total possible ways to select 5 of them.  However, since they must coincide in m places I derived a formula for the number of pairs of n digit binary numbers that coincide in m places:

2^(n-1)C(n,m)

We need a way of determining whether or not there is a set of 5 numbers that are mutually connected pairs.

Unfortunately I am unsure how to continue.  Any hints?

Posted by Jer on 2009-10-21 15:35:24