Formulae for 3 overlapping sets : GMAT Quantitative Section - Page 3
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# Formulae for 3 overlapping sets

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Re: Formulae for 3 overlapping sets [#permalink]

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13 Aug 2012, 06:53
Bunuel wrote:
nifoui wrote:
Can someone confirm whether this formula is true?

Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither

I am getting confused with all different formulas and have seen people adding the 2*(all three) rather than substracting....

I'd advise to understand the concept rather than to memorize formulas. Look at the diagram below:
Attachment:
Union_3sets.gif

FIRST FORMULA
We can write the formula counting the total as: $$Total=A+B+C-(AnB+AnC+BnC)+AnBnC+Neither$$.

When we add three groups A, B, and C some sections are counting more than once. For instance: sections 1, 2, and 3 are counted twice and section 4 thrice. Hence we need to subtract sections 1, 2, and 3 ONCE (to get one) and subtract section 4 TWICE (again to count section 4 only once).

Now, in the formula AnB means intersection of A and B (sections 3 and 4 on the diagram). So when we are subtracting $$AnB$$ (3 an4), $$AnC$$ (1 and 4), and $$BnC$$ (2 and 4) from $$A+B+C$$, we are subtracting sections 1, 2, and 3 ONCE BUT section 4 THREE TIMES (and we need to subtract section 4 only twice), therefor we should add only section 4, which is intersection of A, B and C (AnBnC) again. That is how thee formula $$Total=A+B+C-(AnB+AnC+BnC)+AnBnC+Neither$$ is derived.

SECOND FORMULA
The second formula you are referring to is: $$Total=A+B+C -$${Sum of Exactly 2 groups members} $$- 2*AnBnC + Neither$$. This formula is often written incorrectly on forums as Exactly 2 is no the same as intersection of 2 and can not be written as AnB. Members of exactly (only) A and B is section 3 only (without section 4), so members of A and B only means members of A and B and not C. This is the difference of exactly (only) A and B (which can be written, though not needed for GMAT, as $$AnB-C$$) from A and B (which can be written as $$AnB$$)

Now how this formula is derived?

Again: when we add three groups A, B, and C some sections are counting more than once. For instance: sections 1, 2, and 3 are counted twice and section 4 thrice. Hence we need to subtract sections 1, 2, and 3 ONCE (to get one) and subtract section 4 TWICE (again to count section 4 only once).

When we subtract {Sum of Exactly 2 groups members} from A+B+C we are subtracting sum of sections 1, 2 and 3 once (so that's good) and next we need to subtracr ONLY section 4 ($$AnBnC$$) twice. That's it.

Now, how this concept can be represented in GMAT problem?

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?

Translating:
"are placed on at least one team": members of none =0;
"20 are on the marketing team": M=20;
"30 are on the Sales team": S=30;
"40 are on the Vision team": V=40;
"5 workers are on both the Marketing and Sales teams": MnS=5, note here that some from these 5 can be the members of Vision team as well, MnS is sections 3 an 4 on the diagram (assuming Marketing=A, Sales=B and Vision=C);
"6 workers are on both the Sales and Vision teams": SnV=6 (the same as above sections 2 and 4);
"9 workers are on both the Marketing and Vision teams": MnV=9.
"4 workers are on all three teams": MnSnV=4, section 4.

Question: Total=?

Applying first formula as we have intersections of two groups and not the number of only (exactly) 2 group members.

Total=M+S+V-(MnS+SnV+SnV)+MnSnV+Neither=20+30+40-(5+6+9)+4+0=74.

Example #2:
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?

Translating:
"Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs"
Total=59;
Neither=0 (as members are required to sign up for a minimum of one);
"22 students sign up for the poetry club": P=22;
"27 students for the history club": H=27;
"28 students for the writing club": W=28;
"6 students sign up for exactly two clubs": {Exactly 2 groups members}=6, so sum of sections 1, 2, and 3 is given to be 6, (among these 6 students there are no one who is the member of ALL 3 clubs)

"How many students sign up for all three clubs": question is $$PnHnW=x$$. Or section 4 =?

Apply second formula: $$Total=P+H+W -$${Sum of Exactly 2 groups members}$$-2*PnHnW + Neither$$ --> $$59=22+27+28-6-2*x+0$$ --> $$x=6$$.

Similar problem at: ps-question-94457.html#p728852

Hope it helps.

This was what i was looking for ,instead of solving a complex ven dia every time, these formulas save 10X more time.Thanks and kudos to you.! month away from my test date and i am learning these now. .
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Re: Formulae for 3 overlapping sets [#permalink]

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13 Aug 2012, 19:22
Bunuel

Can you please explain the belovv one too

In a village of hundred households, 75 have at least one DVD player, 80 have at least one cell phone, and 55
have at least one MP3 player. Every village has at least one of these three devices. If X and Y are respectively
the greatest and the lowest possible number of households that have all the three of these devices, X-Y is :
A) 65 B) 55 C) 45 D) 35 E) 25
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Re: Formulae for 3 overlapping sets [#permalink]

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14 Aug 2012, 00:17
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venmic wrote:
Bunuel

Can you please explain the belovv one too

In a village of hundred households, 75 have at least one DVD player, 80 have at least one cell phone, and 55
have at least one MP3 player. Every village has at least one of these three devices. If X and Y are respectively
the greatest and the lowest possible number of households that have all the three of these devices, X-Y is :
A) 65 B) 55 C) 45 D) 35 E) 25

This question is discussed here: in-a-village-of-100-households-75-have-at-least-one-dvd-98257.html

Hope it helps.
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Re: Formulae for 3 overlapping sets [#permalink]

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01 Oct 2012, 16:45
I'm still confused by this. This is the question I'm trying to solve and neither of those formulas give the correct answer.

Quote:
Of the students in Tanner's class, 8 wore a hat to school, 15 students wore gloves, and 10 wore scarves. None of the students wore a scarf without gloves. Four students wore a hat, gloves, and a scarf. Half of the students who wore a hat also wore gloves. How many students are in Tanner's class?

Using the first formula, I get 23. 8+15+10-(14)+4 = 23
Using the 2nd formula, I get 11. 8+15+10-(14)-2(4)=11

Neither of those is correct. The correct answer is 19 according to Kaplan.
Anyone have any insight?

EDIT:
I think I may have figured it out, someone please tell me if this is correct:
Using the first formula:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC) + N
8+15+10-4-4-10+4 = 19
I was missing the 2nd 4 from the Hats & Scarves category since they don't explicitly break it out. So it would be the 4 that are all 3 are also in the A&C group. Is that the correct logic?
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Re: Formulae for 3 overlapping sets [#permalink]

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14 Oct 2013, 15:37
dimri10 wrote:
can som1 please explain why use the 2nd formula in the next 2 cases:
the first demands exactly 2 so it's clear why using the second, but the 2nd question deals with all classes?

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English.
Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four
students are registered for English. If only three students are registered for all three classes, how many students
are registered for exactly two classes?
13 10 9 8 7

Total=A+B+C-[exactly two]-2(all three)+[no group]

68=25+25+34-x-2(3)+0

68=84-x-6

68=78-x

x=10

We use the second formula here, because the first is only for when we take out AnB, AnC, or BnC wholesale. Think of it like a meat cleaver, in the first formula we're chopping out anything that falls into BOTH AnB, even if it also has some C in it. So we do that 3 times, and what you end up doing is needing to add in the AnBnC group again, because you've bludgeoned it repeatedly.

The second formula is when we have something more precise; we are looking for (or are given) the number of items/people/whatever that are EXACTLY in AnB, for instance, with no other interference from group C. So that middle of Bunuel's diagram, that all 3 group, that's unscathed. So we don't need to do any trickery to add it back into the total. In fact, since anything that falls into the (AnBnC) group got counted 3 times (once each when we counted A, B, and C), we need to subtract twice it's value, so that it only gets counted once.

Not sure if that helps, but I was having trouble understanding, and a light flicked on over my head when I thought of it like that.

dimri10 wrote:
4. Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of
three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A
total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing
club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?

In this case again, Total=A+B+C-[exactly two]-2(all three)+[no group]

so

59=22+27+28-6-2x

59=71-2x

x=6

We're using the second formula, again, because we know EXACTLY how many students sign up for EXACTLY two clubs. Note that it's not saying at least two clubs, but exactly two. So they're falling into regions 1, 2, and 3 from Bunuel's chart. Region 4, what you're looking for is completely unknown in this case. Again, hope that helps
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Re: Formulae for 3 overlapping sets [#permalink]

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09 Nov 2013, 05:40
The explanation is excellent. I read this only a week before my test. The explanation along with some of the examples, have really helped.

I used to memorize the formulae for these questions. After going through this post, my understanding has improved (using the venn diagram) tremendously.

I got a question on this on my exam.... Hence, really grateful for the explanation here. Thanks Bunuel....
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Re: Formulae for 3 overlapping sets [#permalink]

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15 Nov 2014, 17:06
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23 Nov 2015, 03:31
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Re: Formulae for 3 overlapping sets [#permalink]

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10 Jul 2016, 03:13
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

A. 65

B. 55

C. 45

D. 35

E. 25

I did understand that X would be 55 but did not understand how y / minima is calculated. Can anyone please explain this again? It isn't clear
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Re: Formulae for 3 overlapping sets [#permalink]

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10 Jul 2016, 04:21
kamkaul wrote:
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

A. 65

B. 55

C. 45

D. 35

E. 25

I did understand that X would be 55 but did not understand how y / minima is calculated. Can anyone please explain this again? It isn't clear

You can find several different solutions of this problem here: in-a-village-of-100-households-75-have-at-least-one-dvd-98257.html
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Re: Formulae for 3 overlapping sets   [#permalink] 10 Jul 2016, 04:21

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# Formulae for 3 overlapping sets

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