Five distinct ndigit 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^(n1)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 20091021 15:35:24 