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There are eight people in a certain club, including Bob and Ted.

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There are eight people in a certain club, including Bob and Ted.  [#permalink]

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New post 27 Mar 2019, 11:30
1
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

61% (01:29) correct 39% (02:04) wrong based on 36 sessions

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There are eight people in a certain club, including Bob and Ted. If the club is to select a 4 people committee that include Bob but not include Ted, how many different such committees are possible?

(A) 20

(B) 28

(C) 30

(D) 35

(E) 56

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Re: There are eight people in a certain club, including Bob and Ted.  [#permalink]

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New post 27 Mar 2019, 12:47
Just remove Bob and Ted from the calculation.
6C3 = 20
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Re: There are eight people in a certain club, including Bob and Ted.  [#permalink]

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New post 29 Mar 2019, 10:32
Using slot method:

We have to include Bob but exclude Ted, so we have to choose 4 people out of 8-1=7 (because Ted is no longer a candidate): Bobx6x5x4=120. Now, since order doesn't matter among the remaining 3 members, we have to exclude their arrangement possibilities as well by dividing 3!: 120/3!=120/6=20

Answer is A.
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Re: There are eight people in a certain club, including Bob and Ted.  [#permalink]

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New post 29 Mar 2019, 11:25
removing Bob and Ted we have to chose 3 people out of 6 and order does not matter
so 6C3 gives us 20 i.e. A
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Re: There are eight people in a certain club, including Bob and Ted.   [#permalink] 29 Mar 2019, 11:25
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There are eight people in a certain club, including Bob and Ted.

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