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Bunuel
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6 x 5 x 4 = 120 = E

6 members. 3 slots. Members can't be selected multiple times.
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Imo it should 6p3... I.e. 120.

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kunalsinghNS
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 !
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Is it 20?
6c3?
how is it 6p3? thanks
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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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Bunuel
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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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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6C3
Choice-A
I consider it below 500 Level Question
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