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A university needs to select a nine-number committee on

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A university needs to select a nine-number committee on [#permalink] New post 30 Jan 2008, 22:07
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A university needs to select a nine-number committee on extracurricular life, whose members must belong to the student government or to the student advisory board. If the student government consists of 10 members, the students advisory board consists of 8 members, and 6 students hold membership in both organizations, how many different committees are possible?

72
110
220
720
1096
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Manager
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Schools: Booth, Stern, Haas
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Re: combs ad counting [#permalink] New post 31 Jan 2008, 03:24
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marcodonzelli wrote:
A university needs to select a nine-number committee on extracurricular life, whose members must belong to the student government or to the student advisory board. If the student government consists of 10 members, the students advisory board consists of 8 members, and 6 students hold membership in both organizations, how many different committees are possible?

72
110
220
720
1096


I think it is C

We have to find how many students we can choose from

10+8-6=12

We have nine places to locate them
12C9=12!/9!3!=220

is it correct?
Director
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Re: combs ad counting [#permalink] New post 31 Jan 2008, 03:52
kazakhb wrote:
what does it mean?


You nailed it!

We have to find out how big our pool of choices is, in this case it's 12 people (10+8-6=12).

Now we know we have 12 eligible people to choose from, since order doesn't matter in our committee we need to find how many combinations of 9 people there are in our group of 12.

12C9 = 220

Answer C
Manager
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Re: combs ad counting [#permalink] New post 31 Jan 2008, 03:59
eschn3am wrote:
kazakhb wrote:
what does it mean?


You nailed it!

We have to find out how big our pool of choices is, in this case it's 12 people (10+8-6=12).

Now we know we have 12 eligible people to choose from, since order doesn't matter in our committee we need to find how many combinations of 9 people there are in our group of 12.

12C9 = 220

Answer C

It is my first right answer in combs and perms;)
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Re: combs ad counting [#permalink] New post 31 Jan 2008, 04:35
kazakhb wrote:
eschn3am wrote:
kazakhb wrote:
what does it mean?


You nailed it!

We have to find out how big our pool of choices is, in this case it's 12 people (10+8-6=12).

Now we know we have 12 eligible people to choose from, since order doesn't matter in our committee we need to find how many combinations of 9 people there are in our group of 12.

12C9 = 220

Answer C

It is my first right answer in combs and perms;)


The first of many I'm sure! Once you begin to understand the logic behind combs, perms and probability they almost become easier than most other problems. Like with most things on the GMAT, practice, practice, practice is key.
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Re: combs ad counting [#permalink] New post 31 Jan 2008, 15:58
marcodonzelli wrote:
A university needs to select a nine-number committee on extracurricular life, whose members must belong to the student government or to the student advisory board. If the student government consists of 10 members, the students advisory board consists of 8 members, and 6 students hold membership in both organizations, how many different committees are possible?

72
110
220
720
1096


I believe we should treat this as a double matrix at first or venn b/c we have some overlap of members here.

10-6 = 4. 8-6=2 So 4+2+6 = 12 total members

12!/9!*3! ---> 220

C
Re: combs ad counting   [#permalink] 31 Jan 2008, 15:58
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