Author 
Message 
TAGS:

Hide Tags

Director
Joined: 30 Jun 2008
Posts: 824

Formulae for 3 overlapping sets
[#permalink]
Show Tags
Updated on: 15 Mar 2019, 02:49
Formulae to solve problems with 3 Overlapping sets u = 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 FOR MORE ON OVERLAPPING SETS CHECK UNDER THE SPOILER
_________________
"You have to find it. No one else can find it for you."  Bjorn Borg
Originally posted by amitdgr on 12 Sep 2008, 03:41.
Last edited by Bunuel on 15 Mar 2019, 02:49, edited 2 times in total.
Updated.




Math Expert
Joined: 02 Sep 2009
Posts: 59182

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
25 May 2010, 06:48
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 240492 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.
_________________




Director
Joined: 05 Jul 2008
Posts: 950

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
29 Sep 2008, 10:43
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.




Intern
Joined: 29 Sep 2008
Posts: 34

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
29 Sep 2008, 14:22
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)



Manager
Joined: 17 Mar 2009
Posts: 193

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
21 Jul 2009, 18:24
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.



Manager
Joined: 29 Oct 2009
Posts: 174

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
31 Oct 2009, 14:35
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?
_________________
Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



Manager
Joined: 04 May 2009
Posts: 113
Location: London
Schools: Haas (WL), Kellogg (matricultating), Stanford (R2, ding), Columbia (ding)
WE 1: 3 years hotel industry sales and marketing France
WE 2: 3 years financial industry marketing UK

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
24 May 2010, 13:21
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....
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 59182

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
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.
_________________



Manager
Joined: 17 Dec 2010
Posts: 87
Location: Australia
GPA: 3.37
WE: Engineering (Consulting)

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
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?
_________________
Kudos always appreciated if my post helped you



Math Expert
Joined: 02 Sep 2009
Posts: 59182

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
13 Feb 2011, 03:41
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.
_________________



Retired Moderator
Joined: 20 Dec 2010
Posts: 1557

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
30 May 2011, 14:48
bblast wrote: source gclub3 Q7. http://gmatclub.com/tests/m03#expl7Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above? 5% 10% 15% 25% 30% Second formula: Total = n(A)+n(B)+n(C)(Sum of Exactly two)2*n(All)+n(None) 100 = 60+50+35SoE22*10+0 SoE2 = 125100=25% Ans: "D"bblast wrote: Query Why does the above logic does not hold in all examples discussed in this forum. I found this alternate method a great shortcut and even applied it to a couple questions.
Yes, you are right!! We can't just apply the formula and get the result in some of these cases. bblast wrote: 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: this formula worked.
Second formula: Total = n(A)+n(B)+n(C)(Sum of Exactly two)2*n(All)+n(None) 85 = 50+30+20SoE22*5 SoE2=9085=5 5% like exactly 2 products. 5% like all 3. 5+5=10% like more than 1 product.
Ans: 10%bblast wrote: 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: this formula worked.
Second formula: 70 = 40+30+35SoE22*15 SoE2 = 5
Ans: 5bblast wrote: 3>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 Answer: could not apply this shortcut
True, I couldn't apply the formula either. Neither am I aware of any formula that can directly solve this. Anyone? I just used some Venn like pictures to understand and get at the result.
Max intersection is easy: 55  Think it as three concentric circles. Innermost circle:55 Intermediate: 75 Outermost: 80
Min intersection was tricky: Make 8020 region as two distinct entity. We need to fit: 55+75=130 in this 8020 disjointed region. We can lay over two products' count on the 20 region: Managed to get rid of 40 that way. Rest is 80 region: We should just layover one product over this 80 region. 13040=90 This is 10 more than 80. Thus, 10 is the minimum intersection within 80 region that we will have to use. Difference: 5510=45
Ans: 45bblast wrote: 4>ALL of the cofee mixtures sold in a certain store contain either a Colombian, Jamaican or Brazilian cofee or some combinations of these. Of all the mixtures 33 contain Colombian cofee, 43 contain Jamaican cofee and 42 contain Brazilian cofee. Of these, 16 contain atleast Colombian and Jamaican cofees, 18 contain atleast Jamaican and Brazilian cofees, 8 contain atleast Brazilian and colombian cofees and 5 contain all three. How many different mixtures are sold in the store ? Answer: this formula failed where it really looked like it would work.
Used top down approach. All=5 C+J=16; Only C+J=165=11 J+B=18; Only J+B=185=13 C+B=8; Only C+B=85=3 C=33; Only C = 33(Only C+J)(Only C+B)All=331135=14 J=43; Only J = 43(Only C+J)(Only J+B)All=4311135=14 B=42; Only B = 42(Only J+B)(Only C+B)All=421335=21
Total = 5+11+13+3+14+14+21=81 Ans: 81bblast wrote: 5>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?
Second formula: 150=66+45+42272*3+None None=150120=30 Ans: 30
_________________



Manager
Joined: 16 May 2011
Posts: 150
Concentration: Finance, Real Estate
GMAT Date: 12272011
WE: Law (Law)

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
07 Jun 2011, 04:52
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. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour 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
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?



Manager
Joined: 16 May 2011
Posts: 150
Concentration: Finance, Real Estate
GMAT Date: 12272011
WE: Law (Law)

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
07 Jun 2011, 05:03
thats what i tried to realize: is it true to say that when u look for ALL or EXACTLY 2 use the 2nd formula and when you look for Neither or Total use the 1st.
in the case of the coffee: 43+42+33(16+18+8) +5 gives 81 [quote][/quote]bblast wrote: 4>ALL of the cofee mixtures sold in a certain store contain either a Colombian, Jamaican or Brazilian cofee or some combinations of these. Of all the mixtures 33 contain Colombian cofee, 43 contain Jamaican cofee and 42 contain Brazilian cofee. Of these, 16 contain atleast Colombian and Jamaican cofees, 18 contain atleast Jamaican and Brazilian cofees, 8 contain atleast Brazilian and colombian cofees and 5 contain all three. How many different mixtures are sold in the store ? Answer: this formula failed where it really looked like it would work.
Used top down approach. All=5 C+J=16; Only C+J=165=11 J+B=18; Only J+B=185=13 C+B=8; Only C+B=85=3 C=33; Only C = 33(Only C+J)(Only C+B)All=331135=14 J=43; Only J = 43(Only C+J)(Only J+B)All=4311135=14 B=42; Only B = 42(Only J+B)(Only C+B)All=421335=21
Total = 5+11+13+3+14+14+21=81



Senior Manager
Joined: 03 Mar 2010
Posts: 341

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
10 Sep 2011, 05:22
Quote: Of the 200 members of an association, each member who speaks German also speaks English and 70 of the members speak only Spanish. If no member speaks all three languages, how many of the members speak two of the three languages.
1. 60 of the members only speak English. 2. 20 of the members do not speak any of the three languages.
This is how Venn Diagram will look like. Attachment:
Screen Shot 20110910 at 3.05.10 PM.png [ 50.52 KiB  Viewed 38920 times ]
3 things to notice: 1) All those who speak German speak English, represented by smaller circle inside bigger circle and denoted as y. 2) Some who speak English also speak Spanish, represented by x. 3) No one speaks all three language and hence only two circles overlap. Given: Total number of people=200 Only Spanish = 70 German and English=y English and Spanish=x To Find x+y. 1) Only English = 60. This means English circle  x  y = 60 Not sufficient to know x+y. 2) 20 speaks neither. Total=200 Hence 20020=180 speaks one of the language. That is English Circle+Spanish Circle= 180 Not sufficient to answer x+y Together, From 2, Total who speaks atleast one language= 180, Given Only Spanish = 70 Hence Only English + y + x = 18070=110 From 1 Only English = 60 x+y+60=110 x+y=11060 = 50 OA C.
_________________
My dad once said to me: Son, nothing succeeds like success.



Manager
Joined: 23 Oct 2011
Posts: 80

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
23 Nov 2011, 09:02
fluke wrote: bblast wrote: 3>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 Answer: could not apply this shortcut
True, I couldn't apply the formula either. Neither am I aware of any formula that can directly solve this. Anyone?I just used some Venn like pictures to understand and get at the result. Max intersection is easy: 55  Think it as three concentric circles. Innermost circle:55 Intermediate: 75 Outermost: 80 Min intersection was tricky: Make 8020 region as two distinct entity. We need to fit: 55+75=130 in this 8020 disjointed region. We can lay over two products' count on the 20 region: Managed to get rid of 40 that way. Rest is 80 region: We should just layover one product over this 80 region. 13040=90 This is 10 more than 80. Thus, 10 is the minimum intersection within 80 region that we will have to use. Difference: 5510=45 We use the second formula that Bunuel mentioned: \(Total=D+C+M \){Sum of Exactly 2 groups members} \( 2*DnCnM + Neither\)"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" > Everyone has a device so: Neither = 0 "In a village of 100 households" > \(T=100\) "75 have at least one DVD player" > \(D=75\) "80 have at least one cell phone" > \(C=80\) "55 have at least one MP3 player" > \(M=55\) "If x and y are respectively the greatest and lowest possible number of households that have all three of these devices" > \(DnCnM= x or y\) So: \(2*DnCnM=D+C+M Total \) {Sum of Exactly 2 groups members} > \(DnCnM=55\) {Sum of Exactly 2 groups members}/2In order for DnCnM to be maximum {Sum of Exactly 2 groups members} has to be minimum. It will be minimum when it is equal with \(0\). Therefore \(x=55\). In order for \(DnCnM\) to be minimum {Sum of Exactly 2 groups members} has to be Maximum. The value of {Sum of Exactly 2 groups members} will depend on \(DnCnM\). In order to be maximum: {Sum of Exactly 2 groups members}+\(DnCnM=100\) Therefore {Sum of Exactly 2 groups members}\(=100DnCnM\) Therefore, \(DnCnM=10 > y=10\) \(xy=45\)



Manager
Joined: 23 Aug 2011
Posts: 63

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
13 Aug 2012, 07:53
Bunuel wrote: 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 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 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 \(AnBC\)) 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. 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. .
_________________
Whatever one does in life is a repetition of what one has done several times in one's life! If my post was worth it, then i deserve kudos



Manager
Joined: 02 Nov 2009
Posts: 95

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
13 Aug 2012, 20:22
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, XY is : A) 65 B) 55 C) 45 D) 35 E) 25



Math Expert
Joined: 02 Sep 2009
Posts: 59182

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
14 Aug 2012, 01:17
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, XY is : A) 65 B) 55 C) 45 D) 35 E) 25 This question is discussed here: inavillageof100households75haveatleastonedvd98257.htmlHope it helps.
_________________



Manager
Joined: 26 Sep 2013
Posts: 182
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
14 Oct 2013, 16:37
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. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour 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+34x2(3)+0 68=84x6 68=78x 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+2862x 59=712x 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



Senior Manager
Joined: 17 Sep 2013
Posts: 343

Re: Formulae for 3 overlapping sets
[#permalink]
Show Tags
09 Nov 2013, 06:40
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....
_________________




Re: Formulae for 3 overlapping sets
[#permalink]
09 Nov 2013, 06:40



Go to page
1 2
Next
[ 23 posts ]



