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a group of astronauts

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by EricLien9122 » Tue Dec 09, 2008 10:25 am
12 has prev. experience.
9 doesn't have experience.

12*9*8=864.

I think this is the answer, not 100% sure.
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by muzali » Tue Dec 09, 2008 10:35 am
1 spot out of 12 for experienced and 2 spots out of 9 for inexperienced=
12* 9C2
=12*9*4
432
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by niraj_a » Tue Dec 09, 2008 12:46 pm
12C1 * 9C2 = 432
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by orel » Tue Dec 09, 2008 1:41 pm
thanks everyone!
OA is 432!
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by Stuart@KaplanGMAT » Tue Dec 09, 2008 1:57 pm
EricLien9122 wrote:12 has prev. experience.
9 doesn't have experience.

12*9*8=864.

I think this is the answer, not 100% sure.
The problem with this approach is that you're double counting the inexperienced possibilities.

Let's call the 9 inexperienced astronauts a, b, c, d, e, f, g, h and i.

By simply multiplying 9*8, you've counted a then b and b then a as two different combinations, when in fact it's the same pair. Since you've doubled the number of options, you need to divide your final answer by 2.
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