This is formula.
N(AUBUC) = N(A) +N(B) +N(C) -N(AiB) -N(BiC)-N(Ci) +N(AiBiC)


20 = 12+9+6-N(AiB) -N(BiC)-N(CiA)+2

20 = 27-N(AiB) -N(BiC)-N(CiA)+2

20 = 29-N(AiB) -N(BiC)-N(CiA)

=> N(AiB) +N(BiC)+N(CiA) = 29 - 20 = 9

9 is the answer, i dont see 9 in the above answers....

kishore wrote:This is formula.
N(AUBUC) = N(A) +N(B) +N(C) -N(AiB) -N(BiC)-N(Ci) +N(AiBiC)

9 is the answer, i dont see 9 in the above answers....

Wrong formula!

I'm not 100% sure on your use of "U" and "i" (I always thought that "U" meant union and "i" meant intersection), but here's a formula you can use written in words:

Total # of objects = (# with characterstic A) + (# with characterstic B) + (# with characterstic C) - (# with ONLY A&B) - (# with ONLY A&C) - (# with ONLY B&C) - 2(# with ALL of A&B&C) + (# with NONE of A/B/C)

Note that the first 3 items on the right side of the equation aren't "# with ONLY characterstic X" - those items include all objects that have that characteristic.

To explain the formula: anyone who's in 2 groups is going to be counted twice, so we need to subtract them once; anyone who's in all 3 groups is going to be counted three times, so we need to subtract them twice.

Applying the formula to this question:

20 = 12 + 9 + 6 - (double movie goers) - 2(2) + 0

20 = 27 - doubles - 4

20 = 23 - doubles

So, the number of people who saw 2 movies is 23 - 20 = 3: choose (D).

Stuart could you explain this

anyone who's in all 3 groups is going to be counted three times, so we need to subtract them twice.

Why are you subtracting those in all 3 only twice instead of three times?

Thanks.

AleksandrM wrote:Stuart could you explain this

anyone who's in all 3 groups is going to be counted three times, so we need to subtract them twice.

Why are you subtracting those in all 3 only twice instead of three times?

Thanks.

If we subtract them 3 times, then we're not counting them at all.

We want to count everyone once. If someone is in just one group, they're fine; if they're in two groups, they've been counted one "extra" time; if they're in three groups, they've been counted two "extra" times.

Stuart, I see you're absolutely right in the logic of the formula. But I've also used throughout my career the formula which Kishore has used above.

Here's a link which talks about this. Please pardon me if we're not supposed to give outside links on the forum.

trying to understand the difference between your formula and our formula.

the term AiB used in your formula (the one given on the site) includes people who watched all three movie C also.

The term uesd by Stuart is specific for only those guys who watched only A and B.

The formula is right. But AIB + BIC + CIA is not what is asked.....

this will include people who watched all 3 movies 3 times .....

Kishore has arrived at 9

the answer will be 9 - 3*(AIBIC) = 9-6 = 3

Perfecto. Both formulas are correct for different purposes. Good to know.

Thanks all.