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A committee of three is to be chosen from six. How many unique committ

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A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 17 Aug 2017, 00:09
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A
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C
D
E

Difficulty:

  15% (low)

Question Stats:

63% (00:17) correct 37% (00:14) wrong based on 90 sessions

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New post 17 Aug 2017, 03:15
6C3 = 20

Hence answer = A

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New post 17 Aug 2017, 08:46
kunalsinghNS wrote:
6C3 = 20

Hence answer = A



is it 20?

6C1 * 5C1 * 4C1 =120

Last edited by rocko911 on 17 Aug 2017, 11:00, edited 1 time in total.

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A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 17 Aug 2017, 09:54
6 x 5 x 4 = 120 = E

6 members. 3 slots. Members can't be selected multiple times.

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Re: A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 17 Aug 2017, 10:52
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Imo it should 6p3... I.e. 120.

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A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 17 Aug 2017, 18:33
rocko911 wrote:
kunalsinghNS wrote:
6C3 = 20

Hence answer = A



is it 20?

6C1 * 5C1 * 4C1 =120



What 6*4*3 does that it eliminates the repitations of a member that is selected in the previous slots in the committee but

6C3 tells all possible unique combinations

OA !

Last edited by kunalsinghNS on 17 Aug 2017, 19:15, edited 1 time in total.

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New post 17 Aug 2017, 18:44
Is it 20?
6c3?
how is it 6p3? thanks

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Re: A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 17 Aug 2017, 18:52
Answer is 20
Example take a committee of 2 to be chosen from 4
A,B,C,D be members
Then AB BC CD AD BD AC are the
Unique committees
That is 4C2=6.
Similarly 6C3 there and it is 20../



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Re: A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 22 Aug 2017, 15:47
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Bunuel wrote:
A committee of three is to be chosen from six. How many unique committees result?

(A) 20
(B) 40
(C) 60
(D) 105
(E) 120


Because the order of selection doesn’t matter, this is a combination problem. The number of ways to select 3 people from 6 is 6C3 = 6!/[3!(6-3)!] = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2) = 20.

Answer: A
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A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 22 Aug 2017, 23:00
Quote:
A committee of three is to be chosen from six. How many unique committees result?

(A) 20
(B) 40
(C) 60
(D) 105
(E) 120


6*5*4*3!/(3!*3!)= 20
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Last edited by Mehed iHasan on 14 Sep 2017, 05:10, edited 2 times in total.

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Re: A committee of three is to be chosen from six. How many unique committ [#permalink]

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New post 24 Aug 2017, 03:34
6C3
Choice-A
I consider it below 500 Level Question

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Re: A committee of three is to be chosen from six. How many unique committ   [#permalink] 24 Aug 2017, 03:34
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