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by adam15 » Sat Dec 05, 2009 3:35 pm
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?

8
13
16
24
27
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by salar_snake » Sun Dec 06, 2009 1:23 am
IMO 13:

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.
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by Stuart@KaplanGMAT » Sun Dec 06, 2009 10:46 am
adam15 wrote: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?

8
13
16
24
27
Here's another way to think about it:

There are 4 people on all 3 boards, let's call them A, B, C and D.

So, to start we have:

Charity 1: ABCD
Charity 2: ABCD
Charity 3: ABCD

That accounts for 4 of the 5 people that the boards have in common with each other, but each board needs 1 more person in common with each other charity. So, we need 1 more person for charities 1/2, 1 more person for charities 1/3 and 1 more person for charities 2/3; let's call them E, F and G.

Now we're up to:

Charity 1: ABCDEF
Charity 2: ABCDEG
Charity 3: ABCDFG

Finally, we need to round out each board with 2 more people to get up to 8, so our completed roster is:

Charity 1: ABCDEFHI
Charity 2: ABCDEGJK
Charity 3: ABCDFGLM

A to M is 13 letters, so we have 13 board members in total.
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