Set theory - three set confusion!! - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

Set theory - three set confusion!!

Expert replies

waltz2salsa: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Tue Sep 28, 2010 9:19 amFollowed by: 1 member; Thanked: 3 times

by waltz2salsa » Mon Oct 18, 2010 10:50 am

i am unsure about when to use the following formulas :

1. True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups) as quoted here and

2. P(AuBuC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC) as quoted in this forum as well as lot other places such as https://gmat-maths.blocked/2008/05/ ... mulas.html.

Please explain the different scenarios when they must be applied.....

Like this post

Source: Beat The GMAT — Problem Solving | Mon Oct 18, 2010 10:50 am

waltz2salsa: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Tue Sep 28, 2010 9:19 amFollowed by: 1 member; Thanked: 3 times

by waltz2salsa » Mon Oct 18, 2010 10:56 am

Now a classic example of this query is the use of the latter formula in the following problem...

Each of 3 charities in Ginglo Kiru has 8 persons serving on its board of directors. if exactly 4 persons serve on 3 board each and each pair of charities has 5 persons in common on their boards of directors, then how many distinct persons serve on one or more boards of directors?
A. 8
B. 13
C. 16
D. 24
E. 27

Solution:

exactly 4 persons serve on 3 board each --> commons (A,B,C) =4

each pair of charities has 5 persons in common --> commons (A,B) = 5, commons (A,C) = 5, commons (B,C) = 5.

distinct persons = n(A) + n(B) + n(C) - n[commons (A,B), commons (A,C), commons (B,C)] + n[commons(A,B,C)] =

---> 8 + 8 + 8 - 5 - 5 - 5 + 4 =24 - 15 + 4 = 13
My questions is that how to identify that here we should use this formula and not the 1st formula

Like this post

GMAT/MBA Expert

Whitney Garner: GMAT Instructor; Posts: 273; Joined: Tue Sep 21, 2010 5:37 amFollowed by: 74 members; Location: Durham, NC; Thanked: 154 times; GMAT Score:770

by Whitney Garner » Mon Oct 18, 2010 12:57 pm

waltz2salsa wrote: My questions is that how to identify that here we should use this formula and not the 1st formula

Both will give the same result if you have complete information about the problem, the question is WHICH will be easier to use based on what you are given. If you know specific information about the overlaps of each group (as in the example you provided above), then the second formula is potentially easier. If you are given information about the totals in each group (those in 3 committees and those in 2 committees and the numbers for each.

First - let's set up a Venn diagram with 3 circles to get the exact numbers from your problem. Since each charity has 8 members, each full circle should contain 8 people. The center overlap (where all 3 charities meet) will contain 4 people since 4 people serve on 3 boards. Then we see that each pair has 5 persons in common on their boards (so given the 4 serving on 3, we need 1 additional person in each paired overlap to give 5 in common between pairs). That leaves 2 people per committee that serve only on that committee. From our Venn we can just count up the sections and find 13 total people divided among the committees. Now we can use this information to re-write the question you presented so that using the first formula would make more sense.

Each of 3 charities in Ginglo Kiru has 8 persons serving on its board of directors. If 4 persons serve on all 3 boards and 3 persons serve on exactly 2 boards each, how many distinct persons serve on one or more boards of directors?

Now we have information that better fits in the first model you provided:

(total# grp1) + (total# grp2) + (total# grp3) - (#exactly 2 grps) - 2(#all 3 grps) = 8 + 8 + 8 - 3 - 2(4) = 13.

So the answer to your original question - the formulas give IDENTICAL solutions, it just depends on the way the information is presented.

Whit

Whitney Garner
GMAT/GRE/EA Instructor & Anxiety/Accommodations Coach
www.whitneygarner.com

Contributor to Beat The GMAT!

Math is a lot like love - a simple idea that can easily get complicated

Like this post

waltz2salsa: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Tue Sep 28, 2010 9:19 amFollowed by: 1 member; Thanked: 3 times

by waltz2salsa » Mon Oct 18, 2010 10:38 pm

Thanks Whitney; this looks like a neat explanation. The challenge for me was not how to solve the problem because using Venn diagram i was easily able to but the problem was to understand the formulas that i had asked. Your solution made things clearer to me. Now correct me if i am wrong but my take away from this is :

#exactly 2 grps = {n[A intersection B]-n[commons(A,B,C)] } + { n[A intersection C] -n[commons(A,B,C)] } +
{ n[C intersection B] -n[commons(A,B,C)] }

and hence this will give us

- #exactly 2 grps - 2(#all 3 grps) = - {n[A intersection B]-n[commons(A,B,C)] } - { n[A intersection C] -n[commons(A,B,C)] } -
{ n[C intersection B] -n[commons(A,B,C)] } - 2 * n[commons(A,B,C)]

= - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)

and hence deriving the second formula from the first formula!!!

Thanks,
Waltz2Salsa

Like this post

GMAT/MBA Expert

Whitney Garner: GMAT Instructor; Posts: 273; Joined: Tue Sep 21, 2010 5:37 amFollowed by: 74 members; Location: Durham, NC; Thanked: 154 times; GMAT Score:770

by Whitney Garner » Tue Oct 19, 2010 3:58 am

waltz2salsa wrote:Thanks Whitney; this looks like a neat explanation. The challenge for me was not how to solve the problem because using Venn diagram i was easily able to but the problem was to understand the formulas that i had asked. Your solution made things clearer to me. Now correct me if i am wrong but my take away from this is :

#exactly 2 grps = {n[A intersection B]-n[commons(A,B,C)] } + { n[A intersection C] -n[commons(A,B,C)] } +
{ n[C intersection B] -n[commons(A,B,C)] }

and hence this will give us

- #exactly 2 grps - 2(#all 3 grps) = - {n[A intersection B]-n[commons(A,B,C)] } - { n[A intersection C] -n[commons(A,B,C)] } -
{ n[C intersection B] -n[commons(A,B,C)] } - 2 * n[commons(A,B,C)]

= - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)

and hence deriving the second formula from the first formula!!!

Thanks,
Waltz2Salsa

Hi!

You can definitely walk away with that derivation - more importantly for the GMAT, however, is that you have a reliable method that consistently provides the correct answer with the least amount of work/time. In that way, I would say that setting up the Venn is the easiest when you have 3 groups (because there is no formula to remember).

So for the vast majority of GMAT test-takers, study the Venn so that you can nail it every time

Whit

- and also be thankful that 3 group questions on the GMAT are a rare occurrence!

Whitney Garner
GMAT/GRE/EA Instructor & Anxiety/Accommodations Coach
www.whitneygarner.com

Contributor to Beat The GMAT!

Math is a lot like love - a simple idea that can easily get complicated

Like this post

pzazz12: Master | Next Rank: 500 Posts; Posts: 155; Joined: Wed Sep 22, 2010 3:54 am; Thanked: 6 times

by pzazz12 » Fri Oct 22, 2010 4:01 am

waltz2salsa wrote:Thanks Whitney; this looks like a neat explanation. The challenge for me was not how to solve the problem because using Venn diagram i was easily able to but the problem was to understand the formulas that i had asked. Your solution made things clearer to me. Now correct me if i am wrong but my take away from this is :

#exactly 2 grps = {n[A intersection B]-n[commons(A,B,C)] } + { n[A intersection C] -n[commons(A,B,C)] } +
{ n[C intersection B] -n[commons(A,B,C)] }

and hence this will give us

- #exactly 2 grps - 2(#all 3 grps) = - {n[A intersection B]-n[commons(A,B,C)] } - { n[A intersection C] -n[commons(A,B,C)] } -
{ n[C intersection B] -n[commons(A,B,C)] } - 2 * n[commons(A,B,C)]

= - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)

and hence deriving the second formula from the first formula!!!

Thanks,
Waltz2Salsa

Thank you guys......

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review