Hi Greg and Everyone else,
I was completely understanding the solution to this problem until you mentioned the formula i.e. A+M+D -(exactly 2) - 2*(all 3). Doesn’t the principle of Inclusion and Exclusion state that one needs to add the Intersection of all 3 sets since it gets added thrice initially first, gets subtracted thrice so hence it needs to be added again ? Screenshot attached for reference.
Source: brilliant.org
It only gets subtracted twice right?
Hi Ganesh,
It is my understanding that since there are 3 (exactly two) regions and all 3 regions contain the (all 3 intersect) region, when we subtract all 3 (exactly two intersect) regions, the (all 3 intersect) region gets subtracted thrice as well.
Hey, similar doubt as yours until I did some googling.
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)
Here
n(A∩B) = d + g
n(B∩C) = f + g
n(C∩A) = e + g
n(A∩B∩C) = g
So when you expand – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C) this term you get
-(d + g) - (f + g) - (e + g) + g which is -d -f -e +2g
When you take it on the other it become -ve hence 2 times (thrice - once)
Hope this helps
The “exactly two” does not include all three, its only when its exactly 2 regions overlapping
Hey Mohitraj,
It does, thank you.
It makes sense now Ganesh, thank you.
Wait you mean -2g right ? since upon expansion it becomes -3g + g.
Yes, it depends if you are taking on the LHS or RHS side but subtracting 2 time if the other side is a positive integer!
