One Response to “Theory of Partitioning – CAT Scan”
yeah, so liked both questions. The first one was less straightforward as compared to the second….
in fact, when a+b+c+d+e=x, then we have [(5-1)+x]Cx ways of partitioning. Now when you have a+b+c+d+…+j+k=2, and a has to take a minimum value of 1, then let’s say a-1=A,
so
A+b+c+d+….+j+k=1, so number of ways of doing this would be (11-1+1)C1 instead of (11-1+2)C2, an error very likely, if you’ve not understood the concept properly.
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yeah, so liked both questions. The first one was less straightforward as compared to the second….
in fact, when a+b+c+d+e=x, then we have [(5-1)+x]Cx ways of partitioning. Now when you have a+b+c+d+…+j+k=2, and a has to take a minimum value of 1, then let’s say a-1=A,
so
A+b+c+d+….+j+k=1, so number of ways of doing this would be (11-1+1)C1 instead of (11-1+2)C2, an error very likely, if you’ve not understood the concept properly.
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By Anirudh on June 8, 2013 at 5:03 PM