A group of 21 astronauts

[thc013]

I would like help solving this problem...

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 1 person in the crew has previous experience in space flight. How many different crew of this type are possible?

A) 432

B) 594

C) 864

D) 1,330

E) 7,980

[kylo]

Total no of ways to select 1 person with experience - 12.
Total no of ways to select remaining 2 persons - 9C2.

hence total ways - 12 x 9C2 = 432.


Thanks!

[RonPurewal]

Total no of ways to select 1 person with experience - 12.
Total no of ways to select remaining 2 persons - 9C2.

hence total ways - 12 x 9C2 = 432.


Thanks!

good.

for those of you who use the "anagram grid" rather than using combinatorial formulas, "9c2" is the expression that comes from the anagram grid: 9! / (2!7!), or 36.

[bhdeepak]

Ron or any other MGMAT staff,

Whats is wrong in this approach....

we need to select 2 people from 9 ...

so the first one can be selected in 9 ways....and the second one in 8 ways....
so total ways is 9*8=72....


this gives the answer as 12 *72=864....

Would really appreciate your opinion on this....Thanks

[rohit801]

deepak,
differece between permutation and combination.

if one has to choose 2 people from 9 people,ask- does the ORDER matter? That is - is AB different from BA. You'll have to see context. Let's say there is a team with 2 players in it. Tehn you only have 1 combination of the team, say, Jack and Daniels. You can' t say I have 2 teams: jack, Daniels and Daniels, Jack. Here the ORDER doeesn't matter. BUT, if they race each other, ORDER matter: Ex jack can come first and Daniels second, and vice-versa[ assuming no tie].

So, if you do it your way then you are overcounting coz you are assuming that order matters, whereas, it doesn't in this case.

BTW: one overcounts exactly n!, where n= number of objects chosen - can u see why?

Hope it helps...

[bhdeepak]

Nicely done...Thanks Rohit

[RonPurewal]

Ron or any other MGMAT staff,

Whats is wrong in this approach....

we need to select 2 people from 9 ...

so the first one can be selected in 9 ways....and the second one in 8 ways....
so total ways is 9*8=72....


this gives the answer as 12 *72=864....

Would really appreciate your opinion on this....Thanks

rohit's post gives a fairly comprehensive summary of the relevant theory here.
here's a specific explanation of why this approach fails in this case:
if you just do 9 x 8, then you are counting each possible SEQUENCE of two people as a separate outcome. for instance, you would be counting "al, bob" and "bob, al" as two separate groups of two people.

notice that if you divide your answer by 2 -- canceling out this double-counting -- then you arrive at the correct answer.

[justinpatellaro]

Can someone please explain how the above astronaut problem is different that this problem:
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
a-certain-law-firm-consists-of-4-senior-partners-t4110.html

Both problems appear the same to me but are solved completely different... When I came across the astronaut problem I began solving it the way the Law office problem is solved, which is not correct... Now Im trying to figure out how to decipher the difference so I dont run into this problem on test day...

[mschwrtz]

The difference is the phrase "at least." In the law firm problem, if we solve directly for the number of acceptable groups (rather than for its complement, the number of unfavorable groups), we'll have to solve for

the number of groups with 3 senior partners and 0 junior partners
plus
the number of groups with 2 senior partners and 1 junior partner
plus
the number of groups with 1 senior partner and 2 junior partner

Each of those can be determined just as you would determine combinations in the astronaut problem.

With an "at least" problem, though, its almost always more efficient to solve for the complement, in this case the number of teams with no senior partners.

Once you've figured out how many different groups of 3 junior partners can be chosen out of a pool of 6, you don't have to ask how many groups of 0 senior partners can be chosen out of a pool of 4.

[justinpatellaro]

Thank you!

[mschwrtz]

my pleasure