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A university needs to select a nine-member committee on extr

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sumana
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A university needs to select a nine-member committee on extr [#permalink] New post   08 Jun 2010, 13:30
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A university needs to select a nine-member committee on extracurricular life, whose members must belong either to the student government or to the student advisory board. If the student government consists of 10 members , the student advisory board consists of 8 members and 6 students hold membership in both organizations , how many different committee are possible.

A. 72
B. 110
C. 220
D. 720
E. 1096
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C
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Re: P&C [#permalink] New post   08 Jun 2010, 14:04
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sumana wrote:
A university needs to select a nine-member committe on extracurricular life, whose members must belong either to the student government or to the student advisory board. If the student government consists of 10 members , the student advisory board consists of 8 members and 6 students hold membership in both organiztions , how many different committes are possible.

a) 72
b) 110
c) 220
d) 720
e) 1096


Two groups:
Student government = 10;
Student advisory board = 8;
Both = 6.

Total # of members of either of group is: 10+8-6=12.

# of ways to choose 9 people out of 12 is \(C^9_{12}=\frac{12!}{9!3!}=220\).

Answer: C.

Hope it helps.
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   15 Feb 2014, 07:50
is the difficulty really sub600?
I find it more difficult.
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   15 Feb 2014, 10:30
b2bt wrote:
is the difficulty really sub600?
I find it more difficult.


I think it is 600-700 level. Compared to OG which are sub 600, this question is actually good.

Combination of two concepts (Venn Diagram and Combination) -> 600-700 level.
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   25 Apr 2015, 11:20
How is 9C12 = 220? I keep getting just 20.
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   25 Apr 2015, 11:35
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How is 9C12 = 220? I keep getting just 20.


12C9 = 12 ! /(9! * 3!) = 10 * 11 * 12/ 2*3 = 10* 11 * 2 = 220.

I hope this is clear.
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   26 Apr 2015, 13:00
Hello,

Step 1:
student government consists of 10 members + the student advisory board consists of 8 members (so add them): 10 + 8 = 18

Step 2:
6 students hold membership in both organizations ( so subtract from total in step 1): 18- 6 = 12

Step 3:
How many different committee are possible if 9 will be selected.
So 12! / 9! 3! (9 will be selected and 3 will not be selected).

12 x 11 x 10 / 3 x 2 x 1 = 220 Answer is C.
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   17 Jul 2016, 19:07
Total number of students = 10+8-6 = 12. Ways to select 9 out of 12 = 12C9 = 12*11*10/3*2=220

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Re: A university needs to select a nine-member committee on extr [#permalink] New post   12 Sep 2016, 11:50
No. of students in government = 10
No. of students in advisory board = 8
No. of students in both = 6
Total = 10 + 8 - 6 = 12

Total number of 9 member committees would be 9C3 = 220.

I wonder if the question could be made a 700 difficulty one by adding constraints. Such as exactly one student should be from the one of the two groups. Food for thought!
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   23 Jan 2018, 09:01
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sumana wrote:
A university needs to select a nine-member committee on extracurricular life, whose members must belong either to the student government or to the student advisory board. If the student government consists of 10 members , the student advisory board consists of 8 members and 6 students hold membership in both organizations , how many different committee are possible.

A. 72
B. 110
C. 220
D. 720
E. 1096


The number of students who belong either to the student government or to the student advisory board or both is:

Total = # Government + # Advisory - # Both + # Neither

10 + 8 - 6 + 0 = 12 students.

So the number of ways to select the 9-member committee out of 12 possible candidates is 12C9:

12!/[(12-9)! x 9!]

12!/(3! x 9!) = (12 x 11 x 10)/3! = (12 x 11 x 10)/(3 x 2) = 2 x 11 x 10 = 220

Answer: C
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   28 Jan 2018, 11:56
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Re: A university needs to select a nine-member committee on extr [#permalink] New post   21 Jul 2023, 01:15
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Re: A university needs to select a nine-member committee on extr [#permalink]
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