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From a group of 21 astronauts that includes 12 people with

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From a group of 21 astronauts that includes 12 people with [#permalink]

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New post Updated on: 25 Jul 2012, 08:07
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From a group of 21 astronauts that includes 12 people with previous experience in space flight, a 3-person crew is to be selected so that exactly one person in the crew has previous experience in space flight. How many different crews of this type are possible?

A. 432
B. 94
C. 864
D. 1330
E. 7980

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Originally posted by Stiv on 25 Jul 2012, 07:49.
Last edited by Bunuel on 25 Jul 2012, 08:07, edited 1 time in total.
Edited the question.
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Re: From a group of 21 astronauts that includes 12 people with [#permalink]

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New post 25 Jul 2012, 08:11
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Stiv wrote:
From a group of 21 astronauts that includes 12 people with previous experience in space flight, a 3-person crew is to be selected so that exactly one person in the crew has previous experience in space flight. How many different crews of this type are possible?

A. 432
B. 94
C. 864
D. 1330
E. 7980


\(C^1_{12}*C^2_9=12*36=432\), where \(C^1_{12}\) is ways to select 1 astronaut with previous experience out of 12 and \(C^2_9\) is ways to select 2 astronauts without experience out of 9 remaining astronauts (21-12=9).

Answer: A.
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Re: From a group of 21 astronauts that includes 12 people with [#permalink]

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New post 15 Oct 2012, 04:19
I apprached this probem as (12 ways to select a previous exp. person)*(9 ways to select 1 person from 9)*(8 ways to select 1 person from remaining 8) ==> 12*9*8 = 864. But from the answer, I see that it has to be divided by 2, but there is no repetition in picking up from 8p after having picked up from 9p.
Can you please let me know what the flaw in my approach is.
I understand the 12C1 * 9C2 approach, wanted to understand this.

Thanks!
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Re: From a group of 21 astronauts that includes 12 people with [#permalink]

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New post 15 Oct 2012, 07:44
rite2mythili wrote:
I apprached this probem as (12 ways to select a previous exp. person)*(9 ways to select 1 person from 9)*(8 ways to select 1 person from remaining 8) ==> 12*9*8 = 864. But from the answer, I see that it has to be divided by 2, but there is no repetition in picking up from 8p after having picked up from 9p.
Can you please let me know what the flaw in my approach is.
I understand the 12C1 * 9C2 approach, wanted to understand this.

Thanks!


(9 ways to select 1 person from 9)*(8 ways to select 1 person from remaining 8)

In this way you count the choices (A,B) and (B,A) separately. The pair A and B should be counted only once, because all that matters, is that both are not experienced, not who they are or in what order they were chosen.
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Re: From a group of 21 astronauts that includes 12 people with [#permalink]

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New post 30 Oct 2014, 23:00
COMB(ENN) where E=experienced and N=no previous experience
12*9*8/(2!). We divide by 2! because there are 2! ways to choose the astronauts without previous experience.
=12*9*4 = 48*9
Because 48*9 ends with a 2, look for the answer that ends with a 2. Only A suffices.
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Re: From a group of 21 astronauts that includes 12 people with [#permalink]

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New post 11 Feb 2018, 22:05
Little confused about this...

I did (12)(9)(8) / 3

12 ways to pick an astronaut, 9 ways to pick without experience, 8 ways to pick without experience....

but I divided by 3 since the order doesn't matter. experience-noexperience-noexperience, noexperience-experience-noexperience, noexperience-noexpereience-experience
Re: From a group of 21 astronauts that includes 12 people with   [#permalink] 11 Feb 2018, 22:05
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