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From a group of 21 astronauts that includes 12 people with previous ex

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Stiv
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From a group of 21 astronauts that includes 12 people with previous ex [#permalink] New post   Updated on: 11 Oct 2019, 05:42
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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. 594
C. 864
D. 1,330
E. 7.980
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A

Originally posted by Stiv on 25 Jul 2012, 07:49.
Last edited by Bunuel on 11 Oct 2019, 05:42, edited 2 times in total.
Edited the question.
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink] 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 previous ex [#permalink] 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 previous ex [#permalink] New post   15 Oct 2012, 07:44
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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 previous ex [#permalink] New post   30 Oct 2014, 23:00
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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 previous ex [#permalink] 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
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink] New post   21 Oct 2018, 13:28
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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

\(21\,\,{\rm{astron}}\,\,\,\left\{ \matrix{\\
\,12\,\,{\rm{experts}} \hfill \cr \\
\,21 - 12 = 9\,\,{\rm{non}}\,{\rm{experts}} \hfill \cr} \right.\)

\(?\,\,\, = \,\,\,\,\# \,\,3\,{\text{people - group}}\,\,,\,\,{\text{exactly}}\,\,1\,\,{\text{expert}}\)


\(?\,\,\, = \,\,\,12 \cdot C\left( {9,2} \right)\,\,\, = 12 \cdot \frac{{9 \cdot 8}}{2} = \underleftrightarrow {9 \cdot 48\,\, = \,\,360 + 72}\,\,\, = \,\,\,432\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink] New post   28 Jan 2019, 02:06
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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



total astronauts w/o exp = 21-12 = 9 out of which 2 are to be chosen ; 9c2
and 1 exp out of 12 ; 12c1
9c2*12c1 ; 432
IMO A
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink] New post   17 Jan 2022, 14:05
Can someone please explain why we could not use this method? Total ways to select 3 from 21 (21c3) - Total ways to select 3 from 9 that have NO experience (9c3) = At least one person with experience.
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink] New post   24 Aug 2022, 21:21
Dufa wrote:
Can someone please explain why we could not use this method? Total ways to select 3 from 21 (21c3) - Total ways to select 3 from 9 that have NO experience (9c3) = At least one person with experience.



This would also include cases where there would be two people with space experience...the question asks for exactly one
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink] New post   25 Aug 2022, 04:54
Total number of astronauts = 21
Number of astronauts with previous work experience = 12
Thus, the number of astronauts with no work experience = 21-12 = 9

In a 3 person crew, the experienced person can be selected in 12 ways.
The other two members can be selected in 9C2 = 9*8/2 = 36

Thus, the total number of ways = 36*12 = 432.

Thus, the correct option is A.
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink] New post   30 Sep 2023, 05:14
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Re: From a group of 21 astronauts that includes 12 people with previous ex [#permalink]
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