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

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Manager
Joined: 16 Feb 2012
Posts: 178
Concentration: Finance, Economics
From a group of 21 astronauts that includes 12 people with  [#permalink]

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Updated on: 25 Jul 2012, 07:07
1
12
00:00

Difficulty:

35% (medium)

Question Stats:

72% (01:07) correct 28% (01:45) wrong based on 406 sessions

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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, 06:49.
Last edited by Bunuel on 25 Jul 2012, 07:07, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 50627
Re: From a group of 21 astronauts that includes 12 people with  [#permalink]

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25 Jul 2012, 07:11
1
2
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).

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Intern
Joined: 23 Aug 2012
Posts: 2
Re: From a group of 21 astronauts that includes 12 people with  [#permalink]

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15 Oct 2012, 03:19
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!
Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: From a group of 21 astronauts that includes 12 people with  [#permalink]

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15 Oct 2012, 06:44
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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30 Oct 2014, 22:00
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.
Intern
Joined: 23 Jul 2015
Posts: 34
Re: From a group of 21 astronauts that includes 12 people with  [#permalink]

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11 Feb 2018, 21:05

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
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 480
Re: From a group of 21 astronauts that includes 12 people with  [#permalink]

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21 Oct 2018, 12:28
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 &nbs [#permalink] 21 Oct 2018, 12:28
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