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2/5 ≤ m/n ≤ 3/5 (Posted on 2009-10-20) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Some Thoughts Stumped | Comment 1 of 7
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
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