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by mkhanna » Thu Sep 24, 2009 8:35 am
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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by bdoug » Thu Sep 24, 2009 9:23 am
We need 3 astronauts on our crew and only one can have experience, meaning the other two members will be inexperienced.

Let's start with the inexperienced members. We have 9 astronauts to choose from (21 astro's - 12 experienced) and we need to choose 2. Using the "choose" function we write this as:

9!/[2! * (9-2)!] or 9!/(2! * 7!)

After some canceling we get (9 * 8)/2! and solve to get 72/2 or 36.

Now we have 12 experienced members to choose from, so we simply multiply the 36 combinations of inexperienced astronauts by 12:

12 * 36 = 432

That's my guess anyway.
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by vivekjaiswal » Thu Sep 24, 2009 7:55 pm
We can select one experienced person from the group of 12 in 12C1 = 12 ways.
We can select the remaining 2 person from the group of 9 inexperienced guys in 9C2 = 9*8/2 = 36 ways
Thus total = 36*12 = 432 (A)

Cheers,
Vivek
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by mkhanna » Sat Sep 26, 2009 7:45 am
Understood. But why don't we do simply 9*8? why are we using 9*8/2?
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by sanjana » Thu Oct 01, 2009 2:29 am
9C2 is

(9!)/(7!x2!) = (9x8x7!)/(7!x2)

=(9x8)/2 = 36.
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by Bullzi » Fri Sep 25, 2015 10:26 am
Hello..

I know this is an old post, still, I had a question I wanted some help with. So, my first instinct on reading this problem was to start with multiplying the 12 with 9 * 8

Now, the only reason this approach isn't right is that the selection of 2 from 9 inexperienced astronauts isn't unique and hence doesn't qualify for the fundamental counting principle to be used?

Thanks
Bullzi
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by Brent@GMATPrepNow » Fri Sep 25, 2015 10:34 am
Bullzi wrote:Hello..

I know this is an old post, still, I had a question I wanted some help with. So, my first instinct on reading this problem was to start with multiplying the 12 with 9 * 8

Now, the only reason this approach isn't right is that selecting 2 from 9 inexperienced astronauts isn't unique and hence doesn't qualify for the fundamental counting principle to be used?

Thanks
Bullzi
We're actually using a mixed approach here (FCP and combinations)
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
Take the task of arranging the 3-person crew and break it into stages.

Stage 1: Select 1 person with previous experience in space flight
There are 12 people to choose from, so we can complete stage 1 in 12 ways

Stage 2: Select 1 person with NO previous experience in space flight
There are 9 people to choose from.
Since the order in which we select the 2 people does not matter, we can use combinations.
We can select 2 people from 9 people in 9C2 ways (36 ways)
So, we can complete stage 1 in 36 ways

If anyone is interested, we have a free video on calculating combinations (like 11C2) in your head: https://www.gmatprepnow.com/module/gmat- ... /video/789

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 3-person crew) in (12)(36) ways ([spoiler]= 432 ways[/spoiler])

Answer: A
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by DavidG@VeritasPrep » Fri Sep 25, 2015 10:37 am
Imagine that instead of choosing 2 astronauts from a group of 9, you were choosing 2 from a group of 3. Call those 3 astronauts A, B, and C. Not hard to just list out all the possible groups of 2: AB, BC, or CA. But note that if we simply did 3*2 = 6, we'd have incorrectly counted AB and BA as unique scenarios. (As well as CA and AC, etc.) Thus, we'd need to do (3*2)/2!, to account for the fact that order does not matter. Similarly if we were choosing 2 from a group of 9, we'd need to (9*8)/2!
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by Matt@VeritasPrep » Fri Sep 25, 2015 10:52 am
Bullzi wrote:Hello..

I know this is an old post, still, I had a question I wanted some help with. So, my first instinct on reading this problem was to start with multiplying the 12 with 9 * 8

Now, the only reason this approach isn't right is that the selection of 2 from 9 inexperienced astronauts isn't unique and hence doesn't qualify for the fundamental counting principle to be used?

Thanks
Bullzi
You're right - the only issue here is that you're doing a PERMUTATION (in which order matters) rather than a COMBINATION (in which order does).

For instance, suppose we have four people and we're picking two. By your logic, we'd have 4 * 3 options, but if we list them out, we see

AB
BA
AC
CA
AD
DA
BC
CB
BD
DB
CD
DC

Notice how we've overcounted? (This is a very easy mistake to make, by the way: I still make it myself sometimes, despite teaching these basic combo problems for five years!)

So we have to cut out the duplicates, and divide by 2. We divide by 2 here because we don't want to arrange our 2 people, so we divide by the number of ways that 2 people can be arranged.
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by Bullzi » Fri Sep 25, 2015 9:15 pm
Thanks all for your responses..

I have another question as a quick follow-up, to check if my understanding so far is right. If the question changes slightly and becomes thus,

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. The 2 remaining persons will be selected for the positions of a Flight Engineer (FE) and a Gravity Analyst (GA). How many different crews of this type are possible?

Based on the changes, I now infer that the selection of the 2 inexperienced astronauts will depend on the order in which they get selected as the selection is for specific positions. A person selected to be an FE isn't the same as him getting selected to be a GA. So, the new solution would be 12 * 9 * 8. Is my inference correct?

Thanks
Bullzi
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by Bullzi » Fri Sep 25, 2015 9:29 pm
COMBINATION (in which order does)
Hi Matt, I assume you meant "in which order doesn't"

Thanks
Bullzi
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by Bullzi » Sun Sep 27, 2015 2:40 am
Checking again.. bump!
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