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Counting problem
By Arun Iyer on Thursday, June 06, 2002 - 07:03 pm:

show that the number of 3-element subsets (a,b,c) of (1,2,3,......,63) with a+b+c<95 is less than the number of those with a+b+c>95

love arun

By Ian Short on Friday, June 07, 2002 - 10:12 am:

If a+b+c < 95 and:

a'=64-a
b'=64-b
c'=64-c

then a'+b'+c' > 97

Is that useful?

By Julian Pulman on Friday, June 07, 2002 - 04:35 pm:

Well, I think what Ian showed is that if we have an instance (a,b,c) that sums to less than 95, then we have also a symmetric instance (64-a, 64-b, 64-c) which sums to greater than 97.
So, if we split the subsets intro two conjugate catagories, each pair being thus {(a,b,c),(64-a,64-b,64-c)}.

The distribution of the subsets is symmetric, which means that the mean equals the median. Because the mean of each pair is 96, the mean of the entire set is 96, and thus the median is also 96.
And as we established, the solution depends on the median being greater than 95, which it is apparently so.

I apologise for any confusion I caused earlier.

Julian

By Ian Short on Saturday, June 08, 2002 - 09:21 am:

Hi, let's say there are precisely N triples whose sum is less than 95. My working shows there are also precisely N triples whose sum is greater than 97 (i.e. triple (a,b,c) corresponding to triple (64-a,64-b,64-c)). That's about as much as I thought when I wrote my original message and it answers the question as 31+32+33=96, so there is at least one more triple with sum greater than 95.

This does essentially establish that the median is 96 (i.e. A+B+C < 96 iff (64-A)+(64-B)+(64-C) > 96) and it's an interesting observation, but it's not necessary to introduce the notion of a median to directly answer the question.

Ian