pankajks2010 wrote:knight247 wrote:
Total number of students=P+H+W-{(P Intersection H)+(P Intersection W)+(H Intersection W)}
-2(P Intersection H Intersection W)+number of ppl in no club
Hi there, Please can you explain why you have taken 2 in the last part of the above mentioned formula??
That is simply a variation of the formula. Read the following link for more info
https://www.urch.com/forums/gmat-problem ... rmula.html
I am abbreviating Intersection with I.Refer to my venn diagram while reading this explanation
From the problem, it is given that no of ppl in exactly two clubs is 6.
i.e. (P I H) + (P I W) + (H I W)=6 .....(1)
Correct? Well, not exactly... because (P I H) + (P I W) + (H I W) contains (P I H I W) thrice coz the intersection of each circle with the subsequent circle contains (P I H I W). Therefore three intersections will contain (P I H I W) thrice. So from the above statement we have to deduct
3(P I H I W). making it
(P I H) + (P I W) + (H I W)-3(P I H I W)=6
(P I H) + (P I W) + (H I W)=6 + 3(P I H I W).....(2)
Plug this into our regular statement of
Total no of items=P+H+W-{(P I H) + (P I W) + (H I W)}+(P I H I W)+Ppl in no group
We Get,
Total no of items=P+H+W-{6 + 3(P I H I W)}+(P I H I W)+Ppl in no group
Total no of items=P+H+W- 6 - 3(P I H I W)+(P I H I W)+Ppl in no group
Total no of items=P+H+W- 6 - 2(P I H I W)+Ppl in no group
Rest is just value subtitution. All clear?
