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Formulae for 3 overlapping sets [#permalink]
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Formulae to solve problems with 3 Overlapping setsu = union and n = intersection1. For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C) 2. To determine the No of persons in exactly one set : P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C) 3. To determine the No of persons in exactly two of the sets : P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C) 4. To determine the No of persons in exactly three of the sets : P(A n B n C) 5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C) (SEE THE ATTACHED PICTURE TO UNDERSTAND THE BELOW MORE CLEARLY) 1. For three sets A, B, and C, P(A u B u C): (A + B + C + X + Y + Z + O) 2. Number of people in exactly one set: ( A+B+C) 3. Number of people in exactly two of the sets: (X+Y+Z) 4. Number of people in exactly three of the sets: O 5. Number of people in two or more sets: ( X+Y+Z+O) 6. Number of people only in set A: A 7. P(A): A+X+Y+O 8. P(A n B): X+O Hope this helps. Please add if i have missed something.
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29 Sep 2008, 10:43
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Great post! How ever, it is very difficult to remember all such formulae, at least for me.
We should be able to derive the rest of the formulae based on
For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
lets discuss how we arrived at each such formula
To determine the No of persons in exactly two of the sets : P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
The reason why we need to subtract 3 times the (A n B n C) is because we need number of persons in exactly 2 sets. How ever, there are people who are in all the three sets and we need to subtract them. In other words when we counted A n B we already counted A n B n C once. Similarly with B n C and C n A. So we subtract them 3 times again.
To determine the No of persons in exactly one set : P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
we need only in one set. that means if we look at the figure we need to find out
only A + only B + only C
Only A = A  { ( A n B) + ( A n C )  (A n B n C) }
The region with only A is the whole region/circle A subtracted from ( A n B) + ( A n C ). In doing so, we accounted for (A n B n C) twice, so subtract (A n B n C) once
Similarly only B and only c
Add them up to get P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
Any takers for the next set of formulas.



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29 Sep 2008, 14:22
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You have done 1, 2, 3: I can do point 4 and 5 in the first list. I am confused what 6 means.
4. To determine the No of persons in exactly three of the sets : this is clearly the people who are a part of all three sets, look at the picture and it is the shared section of all three circles. that is (A n B n C)
5. To determine the No of persons in two or more sets (at least 2 sets) : this is the set of people who are shared by two sets or by all three sets. Let's start with (A n B) + (B n C) + (C n A). But remember, while adding these three, we added the shared part of all three, three times. thus remove that part 2 times (as we want to keep that part too). hence, (A n B) + (B n C) + (C n A)  2( A n B n C)



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seofah wrote: Thanks for putting this thread together, but can we also have some GMAT examples where one can use these concepts? Q1 : In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Answer: 10% like more than 1 product. Q2 : There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German. Answer: 5 students are enrolled in exactly 2 of the courses.



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understudy wrote: This is great stuff. Any idea how to use the above to solve a max / min problem like the following? I've seen some people on here answer it but none of the explanations were that clear.
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
yes.. it is possible to relate it to the formula above. consider the following: Set A  dvd players, Set B  cell phones Set C  mp3 players Now we know that the max value of P (A n B n C) = 55 because C is the limiting factor and its value is 55. Now to find the min value of P (A n B n C), we have: P (A u B u C) = P (A) + P (B) + P (C)  [ P (A n B) + P (A n C) + P (B n C) ] + P (A n B n C) 100 = 75 + 80 + 55  [ P (A n B) + P (A n C) + P (B n C) ] + P (A n B n C) P (A n B n C) = [ P (A n B) + P (A n C) + P (B n C) ]  110 for P (A n B n C) to be min, [ P (A n B) + P (A n C) + P (B n C) ] should be min. this just means that we have to find the min possible overlap between AB, BC and AC and then add them together... how do we do this? simple.. first consider A and B. if the no. of houses having only A = 75 and no. of houses having only B = 80, then the total no of houses having either only A or only B = 155. BUT there are only 100 houses.. therefore a min of 55 houses must have both A and B. another way to understand this is to assume the first 75 houses have A and the next 25 houses have B. this still leaves us with a balance of 55 B. since no house can have more than 1B, a min of 55 houses must have both A and B. Similarly, a min of 35 houses must have both B and C, and a min of 30 houses must have both A and C. therefore, min value of [ P (A n B) + P (A n C) + P (B n C) ] = [55 + 30 + 35] = 120 now substitute this in our above equation and we get min value of P (A n B n C) = 10. thus xy = 5510 = 45 which is choice C. Whats the OA?
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05 Dec 2009, 03:55
crejoc wrote: seofah wrote: Thanks for putting this thread together, but can we also have some GMAT examples where one can use these concepts? Q1 : In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Answer: 10% like more than 1 product. Q2 : There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German. Answer: 5 students are enrolled in exactly 2 of the courses. For Question No. 1 Total = G1+G2+G3  (sum of two products)  2(all three) + neither 85 = 50 + 30 +20  (sum of two products)  2(5) + 0(neither) Let X = Sum of two products therefore, X = 5 For question No. 2 70 = 40+30+352(15)X+0 X = 5
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28 Dec 2009, 13:59
understudy wrote: This is great stuff. Any idea how to use the above to solve a max / min problem like the following? I've seen some people on here answer it but none of the explanations were that clear.
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 it this way... total households = 100 at least 1 dvd player = 75 at least 1 cell phone = 80 at least 1 mp3 player = 55 now households with atleast 1 dvd player and 1 cell phone = (75+80)100= 55 households with atleast 1 dvd player, 1 cell phone and 1 mp3 player = (55+55)100 = 10 so lowest possible number of households with all 3 = 10 max number of house holds with all 3 is equal to the number of households with atleast 1 mp3 player (since it is the smallest in number of the 3) = 55 xy = 55  10 =45 Answer: C can anyone confirm that my method is correct
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Re: Formulae for 3 overlapping sets [#permalink]
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27 Feb 2010, 11:33
meganbaxter1 wrote: I still don't understand the possibility of just deriving all of these in the exam rather than remembering them by heart!
can any one help me with this one:
There are 150 students at Seward High School. 66 students play baseball, 45 play basketball, and 42 play soccer. 27 students play exactly two sports, and three students play all three of the sports. How many of the 150 students play none of the three sports?
A) 0
B) 27
C) 30
D) 99
E) 78 Total Students = GroupA + GroupB +GroupC Exactly Two 2(All Three) + None 150= 66 + 45 + 24 27  2(3) + X X = 30



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Re: Formulae for 3 overlapping sets [#permalink]
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18 Apr 2010, 11:13
Bullet wrote: crejoc wrote: seofah wrote: Thanks for putting this thread together, but can we also have some GMAT examples where one can use these concepts? Q1 : In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Answer: 10% like more than 1 product. Q2 : There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German. Answer: 5 students are enrolled in exactly 2 of the courses. For Question No. 1 Total = G1+G2+G3  (sum of two products)  2(all three) + neither 85 = 50 + 30 +20  (sum of two products)  2(5) + 0(neither) Let X = Sum of two products therefore, X = 5 For question No. 2 70 = 40+30+352(15)X+0 X = 5 For Q1, we need to find those who liked more than 1 product that mean people who liked 2 or 3 products: p(1U2U3) = p(1) + p(2) + p(3)  p(1n2)  p(2n3)  p(1n3) + p (1n2n3) 85=50+30+20 [p(1n2) + p(2n3) + p(1n3)] + p (1n2n3) [p(1n2) + p(2n3) + p(1n3)] = 25. Please post OA and OE.
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Can someone confirm whether this formula is true? Total = Group1 + Group2 + Group3  (sum of 2group 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....
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nifoui wrote: Can someone confirm whether this formula is true? Total = Group1 + Group2 + Group3  (sum of 2group 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 [ 11.63 KiB  Viewed 168433 times ]
FIRST FORMULAWe 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 FORMULAThe 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 \(AnBC\)) 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}\(2*PnHnW + Neither\) > \(59=22+27+2862*x+0\) > \(x=6\). Answer: 6. Similar problem at: psquestion94457.html#p728852Hope it helps.
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25 May 2010, 12:47
Thanks Bunuel for your very clear explanation, this is very helpful, as always!
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04 Jun 2010, 18:48
AWESOME POST Bunuel....I cant thankyou enough



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24 Jul 2010, 23:33
Thanks Bunuel for your very clear explanation.
All your explanations/tips are really very helpful.
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02 Sep 2010, 11:19
harshamadabal wrote: 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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02 Sep 2010, 20:04
its clear now ...thanks Bunuel



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Nice explaination Thanks



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02 Feb 2011, 01:48
Bunuel wrote: I'd advise to understand the concept rather than to memorize formulas. Look at the diagram below: Attachment: Union_3sets.gif FIRST FORMULAWe can write the formula counting the total as: \(Total=A+B+C(AnB+AnC+BnC)+AnBnC+Neither\). ... +Kudo. very clear explanation .



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13 Feb 2011, 00:01
I've got a curly one  which just isn't working out using the formulas (I'm sure I'm doing something wrong though!) Here's the question: The 38 movies in the video store fall into the following 3 categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action & drama, 3 are both action & comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?Solving it via firstprinciples gives me the correct answer  but it takes me something like 4 mins  which is way too slow. Using the formula, I can solve these ones in just under 90 seconds....but this one just plain isn't working out with me: For this question, I am using the following info: A = 10, B = 18, C = 20 Sum of exactly 2 groups = 5 + 3 + 4 = 12 x = ? T = 38 By using the formula: 38 = 10 + 18 + 20  (12)  2x 38 = 48  12  2x 38 = 36  2x x = 1 ????? Any help?
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MackyCee wrote: I've got a curly one  which just isn't working out using the formulas (I'm sure I'm doing something wrong though!)
Here's the question:
The 38 movies in the video store fall into the following 3 categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action & drama, 3 are both action & comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?
Solving it via firstprinciples gives me the correct answer  but it takes me something like 4 mins  which is way too slow. Using the formula, I can solve these ones in just under 90 seconds....but this one just plain isn't working out with me:
For this question, I am using the following info: A = 10, B = 18, C = 20 Sum of exactly 2 groups = 5 + 3 + 4 = 12 x = ? T = 38
By using the formula:
38 = 10 + 18 + 20  (12)  2x 38 = 48  12  2x 38 = 36  2x x = 1
?????
Any help? There is a reason for two formulas in my post. 5 are both action and drama DOES NOT mean that 5 movies are ONLY action and drama, among these 5 might be some which are comedy. Similarly for: "3 are both action and comedy, and 4 are both drama and comedy" So you should apply first formula: {Total} = {action} + {drama} + {comedy}  {action&drama+action&comedy+drama&comedy} + {action&drama&comedy} > 38=10+20+18(5+3+4)+{action&drama&comedy} > {action&drama&comedy}=2. You can apply second formula as well but you should make necessary modification: let {action&drama&comedy}=x, then: ONLY action and drama = (5x); ONLY action and comedy = (3x); ONLY drama and comedy = (4x); {Total} = {action} + {drama} + {comedy}  {EXACTLY two movies}  2*{action&drama&comedy} > 38=10+20+18(5x+3x+4x)2x > x=2. Hope it's clear.
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Re: Formulae for 3 overlapping sets
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13 Feb 2011, 03:41



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