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kamkaul
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Formulae for 3 overlapping sets 10 Jul 2016, 04:13
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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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Formulae for 3 overlapping sets 10 Jul 2016, 05:21
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kamkaul
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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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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Formulae for 3 overlapping sets 06 Jun 2021, 02:13
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Bunuel
nifoui
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 cannot 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.

Answer: 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}\(-2*PnHnW + Neither\) --> \(59=22+27+28-6-2*x+0\) --> \(x=6\).

Answer: 6.

Similar problem at: https://gmatclub.com/forum/ps-question-9 ... ml#p728852

Hope it helps.
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hi VeritasKarishma

can you pls explain the following

i have a question about this formula \(Total=A+B+C -\){Sum of Exactly 2 groups members} \(- 2*AnBnC + Neither\).

does a regions (1, 2 and 3) of exactly 2 group overlaps include those in three group overlap as well ? if so then why are we subtracting three group overlap twice ?

on the other hand if regions (1, 2 and 3) of exactly 2 group overlaps DO NOT include people of three group overlap, then why are we subtracting three group overlap twice, ... if regions of exactly 2 group overlaps DO NOT include people of three group overlap then we have to subract it only once :?
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Formulae for 3 overlapping sets Updated on: 21 Dec 2023, 08:19
Originally posted by KarishmaB on 07 Jun 2021, 00:44.
Last edited by KarishmaB on 21 Dec 2023, 08:19, edited 1 time in total.
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dave13


can you pls explain the following

i have a question about this formula \(Total=A+B+C -\){Sum of Exactly 2 groups members} \(- 2*AnBnC + Neither\).

does a regions (1, 2 and 3) of exactly 2 group overlaps include those in three group overlap as well ? if so then why are we subtracting three group overlap twice ?

on the other hand if regions (1, 2 and 3) of exactly 2 group overlaps DO NOT include people of three group overlap, then why are we subtracting three group overlap twice, ... if regions of exactly 2 group overlaps DO NOT include people of three group overlap then we have to subract it only once :?
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A + B + C = Exactly 1 + Exactly 2 counted twice + Exactly 3 counted thrice

Check this post: https://anaprep.com/sets-statistics-thr ... ping-sets/
It will show with a diagram the various regions.

Note that
A + B + C = (a + d + g + e) + (b + d + g + f) + (c + e + g + f)
Hence, exactly 1 is counted once.
Exactly 2 (d, e, f) are counted twice
Exactly 3 (g) is counted 3 times.

So we subtract "Exactly 2" once and "Exactly 3" two times.
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Formulae for 3 overlapping sets 08 Jan 2024, 07:43
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Dear Bunuel
Is there any formula to calculate min or max of intersection for 3 sets? (If so, would you pls write it?)
I have problem in answering min max in set questions? Is there any source on GMAT Club for it?
Best Regards

Bunuel
harshamadabal
Hi All,

Am a new chap here....

I did not quite understand the below sentence
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)

i understood that section 4 is deleted thrice, but why should it be deleted twice..it should be deleted only once right?

Hi, and welcome to Gmat Club!

Formula is: \(Total=A+B+C-(AnB+AnC+BnC)+AnBnC+Neither\).

Not to have double counting of section 4 we should count it only once. A+B+C counts it 3 times (so to count section 4 once we should subtract it twice), but then when we subtract (AnB+AnC+BnC) we subtract section 4 three times so at this stage we haven't counted it at all, so we should add it once and that' why we have +AnBnC part.

Hope it's clear.
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Formulae for 3 overlapping sets 08 Jan 2024, 08:25
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nazii
Dear Bunuel
Is there any formula to calculate min or max of intersection for 3 sets? (If so, would you pls write it?)
I have problem in answering min max in set questions? Is there any source on GMAT Club for it?
Best Regards

Show more

I know your query is directed toward Bunuel and I am sure he will share some wonderful resources (as always) with you but meanwhile, I am sharing a few of my videos on max min concept in overlapping sets that my learners found very helpful:

How to deal with Max Min in Overlapping Sets: https://youtu.be/oLKbIyb1ZrI
Overlapping Sets Max Min with Atleast One: https://youtu.be/4IujMV4et-U
Overlapping Sets - Max Min given Both and Neither: https://youtu.be/M3wJ15ZC1pY
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Formulae for 3 overlapping sets 08 Jan 2024, 08:31
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Dear karishma
I do appreciate for your videos, they are very useful, I already watched them
still, I am looking for a formula, if it exists, that I can solve such problems fast?
If you know the formulas I would appreciate if you share it
Best regards
KarishmaB
nazii
Dear Bunuel
Is there any formula to calculate min or max of intersection for 3 sets? (If so, would you pls write it?)
I have problem in answering min max in set questions? Is there any source on GMAT Club for it?
Best Regards


I know your query is directed toward Bunuel and I am sure he will share some wonderful resources (as always) with you but meanwhile, I am sharing a few of my videos on max min concept in overlapping sets that my learners found very helpful:

How to deal with Max Min in Overlapping Sets: https://youtu.be/oLKbIyb1ZrI
Overlapping Sets Max Min with Atleast One: https://youtu.be/4IujMV4et-U
Overlapping Sets - Max Min given Both and Neither: https://youtu.be/M3wJ15ZC1pY
Show more
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Formulae for 3 overlapping sets 08 Jan 2024, 08:33
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nazii
Dear Bunuel
Is there any formula to calculate min or max of intersection for 3 sets? (If so, would you pls write it?)
I have problem in answering min max in set questions? Is there any source on GMAT Club for it?
Best Regards

Bunuel
harshamadabal
Hi All,

Am a new chap here....

I did not quite understand the below sentence
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)

i understood that section 4 is deleted thrice, but why should it be deleted twice..it should be deleted only once right?

Hi, and welcome to Gmat Club!

Formula is: \(Total=A+B+C-(AnB+AnC+BnC)+AnBnC+Neither\).

Not to have double counting of section 4 we should count it only once. A+B+C counts it 3 times (so to count section 4 once we should subtract it twice), but then when we subtract (AnB+AnC+BnC) we subtract section 4 three times so at this stage we haven't counted it at all, so we should add it once and that' why we have +AnBnC part.

Hope it's clear.
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I'm not aware of any specific formula for this. However, below are some links that might be useful to you

14. Min/Max Problems



For more check:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
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Formulae for 3 overlapping sets 08 Jan 2024, 23:41
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nazii
Dear karishma
I do appreciate for your videos, they are very useful, I already watched them
still, I am looking for a formula, if it exists, that I can solve such problems fast?
If you know the formulas I would appreciate if you share it
Best regards
KarishmaB
nazii
Dear Bunuel
Is there any formula to calculate min or max of intersection for 3 sets? (If so, would you pls write it?)
I have problem in answering min max in set questions? Is there any source on GMAT Club for it?
Best Regards


I know your query is directed toward Bunuel and I am sure he will share some wonderful resources (as always) with you but meanwhile, I am sharing a few of my videos on max min concept in overlapping sets that my learners found very helpful:

How to deal with Max Min in Overlapping Sets: https://youtu.be/oLKbIyb1ZrI
Overlapping Sets Max Min with Atleast One: https://youtu.be/4IujMV4et-U
Overlapping Sets - Max Min given Both and Neither: https://youtu.be/M3wJ15ZC1pY
Show more


There are no other formulae for max min. The formulae are the same as for regular sets questions. The important point is that to maximise overlap, you would need to maximise neither i.e. bring the circles closer together. The videos discuss the process to follow and the reason behind it and you can very easily solve most min max questions using this process.
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Formulae for 3 overlapping sets 21 Jul 2024, 22:10
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­What if it's 4 circles. Will the method be the same? with only little change to count the 4 overlaps area for 3 times?­
What if question has more than 4? Can a table be made instead to avoid confusion?­
I'm always confused the union circles with mixture table.... I guess only the circles can solve union problems NOT the table which is for mixture problem... right guys?
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Formulae for 3 overlapping sets 21 Jul 2024, 22:49
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sanebeyondone
­What if it's 4 circles. Will the method be the same? with only little change to count the 4 overlaps area for 3 times?­
What if question has more than 4? Can a table be made instead to avoid confusion?­
I'm always confused the union circles with mixture table.... I guess only the circles can solve union problems NOT the table which is for mixture problem... right guys?
Show more
­
I have discussed max-min in 4 overlapping sets here:
https://youtu.be/HzggG6Jd2tU

It doesn't matter how many sets you have.
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Formulae for 3 overlapping sets 14 Dec 2024, 03:47
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Great explanation! Thank you so much!
Bunuel
nifoui
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 cannot 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.

Answer: 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}\(-2*PnHnW + Neither\) --> \(59=22+27+28-6-2*x+0\) --> \(x=6\).

Answer: 6.

Similar problem at: https://gmatclub.com/forum/ps-question- ... ml#p728852

Hope it helps.
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Formulae for 3 overlapping sets 27 Apr 2025, 10:54
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If I've got a 3 set diagram and I want to calculate P(A only), What's the correct formula definition?
P(A Only) = P(A) - P(AnC) - P(AnB) - P(AnBnC)
Or something else?
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Formulae for 3 overlapping sets 27 Apr 2025, 10:56
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etaina
If I've got a 3 set diagram and I want to calculate P(A only), What's the correct formula definition?
P(A Only) = P(A) - P(AnC) - P(AnB) - P(AnBnC)
Or something else?
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Yes, you are almost right.

The correct formula for P(A only) is:

P(A only) = P(A) - P(A∩B) - P(A∩C) + P(A∩B∩C)

You subtract the overlaps with B and C, but since the triple overlap P(A∩B∩C) was subtracted twice (once in P(A∩B) and once in P(A∩C)), you need to add it back once.

So it’s not just subtracting everything, you must add back P(A∩B∩C).
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Formulae for 3 overlapping sets 27 Apr 2025, 11:24
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But is the Intersection BnC not donating an additional AnBnC? since we're in a 3-set problem space
Bunuel
etaina
If I've got a 3 set diagram and I want to calculate P(A only), What's the correct formula definition?
P(A Only) = P(A) - P(AnC) - P(AnB) - P(AnBnC)
Or something else?

Yes, you are almost right.

The correct formula for P(A only) is:

P(A only) = P(A) - P(A∩B) - P(A∩C) + P(A∩B∩C)

You subtract the overlaps with B and C, but since the triple overlap P(A∩B∩C) was subtracted twice (once in P(A∩B) and once in P(A∩C)), you need to add it back once.

So it’s not just subtracting everything, you must add back P(A∩B∩C).
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Bunuel
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Formulae for 3 overlapping sets 27 Apr 2025, 11:39
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etaina
But is the Intersection BnC not donating an additional AnBnC? since we're in a 3-set problem space
Bunuel
etaina
If I've got a 3 set diagram and I want to calculate P(A only), What's the correct formula definition?
P(A Only) = P(A) - P(AnC) - P(AnB) - P(AnBnC)
Or something else?

Yes, you are almost right.

The correct formula for P(A only) is:

P(A only) = P(A) - P(A∩B) - P(A∩C) + P(A∩B∩C)

You subtract the overlaps with B and C, but since the triple overlap P(A∩B∩C) was subtracted twice (once in P(A∩B) and once in P(A∩C)), you need to add it back once.

So it’s not just subtracting everything, you must add back P(A∩B∩C).
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Yes, P(B∩C) includes P(A∩B∩C). But how does this matter?

We subtract P(A∩B) (which means subtracting both P(A and B only) and P(A and B and C)) and P(A∩C) (which means subtracting both P(A and C only) and P(A and B and C). Thus, we subtract P(A∩B∩C) twice. But we need to subtract it only once, so we adjust by adding P(A∩B∩C) once at the end.

The correct formula is:

P(A only) = P(A) - P(A∩B) - P(A∩C) + P(A∩B∩C).

Your worry about P(B∩C) causing issues here is unnecessary. Check the diagram it should become easier to understand.
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Bunuel
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