Re: ISI 2015 PEA Answer Key - Discuss

Posted by onionknight on Apr 14, 2015; 10:20am
URL: http://discussion-forum.276.s1.nabble.com/ISI-2015-PEA-Answer-Key-Discuss-tp7596123p7596167.html

Let the scores of the three people be X1 , X2 and X3. X1, X2 and X3 can take values between 1 and 6. Now you need to find all cases where the sum of these values is 10. X1+X2+X3=10 . So essentially you need to find in how many ways you can express 10 as a sum of 3 integers.

To do these sort of problems, there is a nifty trick. Imagine 10 blocks and 2 dividers, now you have 10+2=12 spaces for these 12 things. When you select any two spaces for the two dividers by 12C2, you essentially divide the 10 blocks into 3 different groups which are your 3 different integers that sum up to 10. But here since X1 and X2 and X3 can't be 0, define Yi=Xi+1 so your problem would reduce to Y1+Y2+Y3=7 which gives (7+2)C2=9C2=36. Now subtract the cases where one of the Yi's is 6 and 7 (since this would make Xi 7 or 8). There will be 6 cases for one of the Yi's=6 and 3 for one of the Yi's=7.
So finally 36-6-3= 27.