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how to solve this

by anjaligeorge1 » Thu Jun 19, 2008 10:41 am
In a survey, 20 students were each asked whether they watched movie A, B or C. According to the survey, every student had watched at least one of the movies, 12 students had watched movie A, 9 students had watched movie B, 6 students had watched movie C, and 2 students had watched all of the three movies. How many of the students surveyed had watched exactly 2 of the movies? 

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
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by kishore » Thu Jun 19, 2008 12:04 pm
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....
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by Stuart@KaplanGMAT » Thu Jun 19, 2008 12:14 pm
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).
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by AleksandrM » Thu Jun 19, 2008 5:09 pm
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.
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by Stuart@KaplanGMAT » Thu Jun 19, 2008 5:51 pm
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.
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by g_beatthegmat » Thu Jun 19, 2008 11:13 pm
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.
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by durgesh79 » Thu Jun 19, 2008 11:33 pm
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
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by g_beatthegmat » Thu Jun 19, 2008 11:56 pm
Perfecto. Both formulas are correct for different purposes. Good to know.

Thanks all.
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