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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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.
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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vivekjaiswal
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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
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
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
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
GMAT/MBA Expert
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Brent@GMATPrepNow
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We're actually using a mixed approach here (FCP and combinations)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
Take the task of arranging the 3-person crew and break it into stages.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
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
--------------------------
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DavidG@VeritasPrep
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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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Matt@VeritasPrep
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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).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
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.
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
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
