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Sourav700
There are two formulas for 3 overlapping sets:
Total=A+B+C−(sum of 2−group overlaps)+(all three)+NeitherTotal=A+B+C−(sum of 2−group overlaps)+(all three)+Neither.

Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+NeitherTotal=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither.



Alright, I'm really really confused on this topic. I'd really appreciate if someone could help me understand the difference between the two.

Bunuel VeritasKarishma IanStewart chetan2u

Let us look at both the formulas..

A=a+d+e+f
B=b+d+f+g
C=c+e+f+g

Main difference is the definition and contents of sum of 2−group overlaps and sum of EXACTLY 2−group overlaps
sum of 2−group overlaps means (d+f)+(e+f)+(g+f) and sum of EXACTLY 2−group overlaps means d+e+f

(I) Total=A+B+C−(sum of 2−group overlaps)+(all three)+Neither
Here sum of 2−group overlaps are (d+f)+(e+f)+(g+f).. So, you can see the (all three, that is f, is getting added thrice , the net result being 0, so you add one back .
all three = f
Neither =0
So Total=A+B+C−(sum of 2−group overlaps)+(all three)+Neither=A=a+d+e+f+b+d+f+g+c+e+f+g-((d+f)+(e+f)+(g+f))+(f)

(II) Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither
sum of EXACTLY 2−group overlaps means d+e+f..
so you are not subtracting all three even once, and they have been added thrice in A+B+C, so subtract it 2 times
Attachments

overlap.png
overlap.png [ 45.86 KiB | Viewed 2371 times ]

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Hi chetan2u

Thanks very much for the explanation. Appreciate it.

i'd like to follow up with another question here.

Now i get the difference between the two:
(sum of 2−group overlaps) = (sum of exactly 2) + 3f. correct?

My question is:

when I'm provided with individual values of d, e, f, and g. why do I get an incorrect answer when i use the second formula?

For example:

Example 1: Workers are grouped by their areas of expertise, and are placed on at least one team. 20 are on the marketing team, 30 are on the Sales team, and 40 are on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers are on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all three teams. How many workers are there in total?

The answer here is 74 and not 62.

1st equation: 20 + 30 + 40 - ..................

wait, i think i got it.

When we are given the value for d e and g we must subtract the value of f from each of these values for them to be valid in the second equation.

1st equation: 30 + 40 + 20 - (5+9+6) + 4 = 74
2nd equation: 30 + 40+ 20 - (1+5+2) - 2(4) = 74

Thanks a lot chetan2u :)

P.S. If you were in Calcutta, i'd buy you pizza :p
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Sourav700
Hi chetan2u

Thanks very much for the explanation. Appreciate it.

i'd like to follow up with another question here.

Now i get the difference between the two:
(sum of 2−group overlaps) = (sum of exactly 2) + 3f. correct?

My question is:

when I'm provided with individual values of d, e, f, and g. why do I get an incorrect answer when i use the second formula?

For example:

Example 1: Workers are grouped by their areas of expertise, and are placed on at least one team. 20 are on the marketing team, 30 are on the Sales team, and 40 are on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers are on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all three teams. How many workers are there in total?

The answer here is 74 and not 62.

1st equation: 20 + 30 + 40 - ..................

wait, i think i got it.

When we are given the value for d e and g we must subtract the value of f from each of these values for them to be valid in the second equation.

1st equation: 30 + 40 + 20 - (5+9+6) + 4 = 74
2nd equation: 30 + 40+ 20 - (1+5+2) - 2(4) = 74

Thanks a lot chetan2u :)

P.S. If you were in Calcutta, i'd buy you pizza :p


Yes, you are correct..
what is given is d+f as 5, so you have to subtract f from it.

I think I can let that Pizza go, even if it means shifting to calcutta. ;)
All the best
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Sourav700
There are two formulas for 3 overlapping sets:
Total=A+B+C−(sum of 2−group overlaps)+(all three)+NeitherTotal=A+B+C−(sum of 2−group overlaps)+(all three)+Neither.

Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+NeitherTotal=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither.



Alright, I'm really really confused on this topic. I'd really appreciate if someone could help me understand the difference between the two.

Bunuel VeritasKarishma IanStewart chetan2u

Check below.

19. Overlapping Sets


[*]Theory
Overlapping Sets Made Easy!
How to draw a Venn Diagram for problems
ADVANCED OVERLAPPING SETS PROBLEMS
Formulae for 3 overlapping sets
[*]Questions
The E-GMAT Sets Triad: 3 Exciting Sets Questions!
The Word “Or” in GMAT Math
DS Questions
PS Questions
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Sourav700

Alright, I'm really really confused on this topic. I'd really appreciate if someone could help me understand the difference between the two.

I've never used those formulas once on a GMAT question, and in my experience, a lot of test takers find them confusing, and often use the wrong one, so they aren't very reliable. Even when using a formula is an option, these overlapping set problems tend almost always to be easier and faster to solve just by using a Venn diagram anyway, as Paresh does on the first page of this thread, and since the Venn diagram can easily be adapted to situations where the formulas don't apply (the types of situations you encounter a lot on the GMAT), it's the better method to learn in general.
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piyushpachori
Ans. A - 62

I solved this using Venn Diagrams. Attaching the solution for you. This is my first ever attachment in the forum. Not sure if it will display the image of the attachment has to be opened. Please check.

Oh no, It's wrong!!

29+2+23+5+4+1+10=74

29 instead of 2
23 instead of 15
10 instead of 21
5 instead of 9
1 instead of 6
2 instead of 5
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I subtracted 2 x 4 rather than adding the 4 giving me "A". From you have listed here you can use either or. Which do you use when? I've never even seen the first one. Simple question aside from a little caveat to trick folks.
Bunuel


There are two formulas for 3 overlapping sets:
\(Total = A + B + C - (sum \ of \ 2-group \ overlaps) + (all \ three) + Neither\).

\(Total = A + B + C - (sum \ of \ EXACTLY \ 2-group \ overlaps) - 2*(all \ three) + Neither\).

For more check here: ADVANCED OVERLAPPING SETS PROBLEMS
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tomloveless
I subtracted 2 x 4 rather than adding the 4 giving me "A". From you have listed here you can use either or. Which do you use when? I've never even seen the first one. Simple question aside from a little caveat to trick folks.

Please check this: ADVANCED OVERLAPPING SETS PROBLEMS
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