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Set theory - three set confusion!!

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Set theory - three set confusion!!

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.....
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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
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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.

:D
Whit
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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
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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! :D
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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......
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