Re: ISI 2015 PEA Answer Key - Discuss

Posted by Rajat on Apr 14, 2015; 10:42am
URL: http://discussion-forum.276.s1.nabble.com/ISI-2015-PEA-Answer-Key-Discuss-tp7596123p7596168.html

Hi Onionknight,

You are right. I am aware of the blocks-and-dividers technique and that is how I have solved it too but i am not able to arrive at 27. I get 33 as the answer. Please tell me where i am going wrong.

There are 10 blocks and two dividers are to be placed in the 9 spaces in between. For the first divider, we have 9 options. For the second divider we have 8 options. So total number of ways is 9x8/2 = 36 unique ways. (Dividing by 2 because there will be duplicate counting). But this 36 number of ways will also include those configurations where one divider is placed in the 7th and 8th space.

Case 1: Now, if one divider is in the 7th space (i.e after the 7th block), the other divider can either be in the 8th space or the 9th space. This means two configurations.
Case 2: If one divider is in the 8th space, the other divider can only be place in the 9th space.This implies one configuration.

So we should subtract these 2+1 impossible configurations from 36 to give 33 .
I know i am wrong somewhere, not sure where !