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

Question

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

A. 432
B. 594
C. 864
D. 1,330
E. 7.980

Bunuel · Accepted Answer

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.

EvaJager · Answer

(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.

rite2mythili · Answer

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!

TooLong150 · Answer

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.

jasonfodor · Answer

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

fskilnik · Answer

21\,\,{m{astron}}\,\,\,\left\{ \matrix{
 \,12\,\,{m{experts}} \hfill \cr 
 \,21 - 12 = 9\,\,{m{non}}\,{m{experts}} \hfill \cr} ight.

?\,\,\, = \,\,\,\,\# \,\,3\,{	ext{people - group}}\,\,,\,\,{	ext{exactly}}\,\,1\,\,{	ext{expert}}

?\,\,\, = \,\,\,12 \cdot C\left( {9,2} ight)\,\,\, = 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.

Archit3110 · Answer

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

Dufa · Answer

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.

abraar308 · Answer

This would also include cases where there would be two people with space experience...the question asks for exactly one

GmatPoint · Answer

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

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