FOR EACH OF THE 12 EXPERIENCED GUYS WE CAN SELECT 2 INEXPERIENCED GUYS FROM THE GROUP OF 9
HENCE IT SHOULD BE 12*(9C2) == 432
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Source: Beat The GMAT — Data Sufficiency |
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vinod_ece66
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VP_Jim
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Allow me to chime in on this one.
I always like to think of permutation/combination problems logically.
First, we have to pick someone with experience, so there are 12 choices.
Next, we have to pick two people who are inexperienced, so there are 9 choices for the first and 8 choices for the second. However, it doesn't matter in which order we pick these people (i.e., picking Al and then Bob is the same thing as picking Bob and then Al). So, we have to divide by 2!.
So we get: 12 * ((9*8)/2!) = 432.
Hope this helps!
I always like to think of permutation/combination problems logically.
First, we have to pick someone with experience, so there are 12 choices.
Next, we have to pick two people who are inexperienced, so there are 9 choices for the first and 8 choices for the second. However, it doesn't matter in which order we pick these people (i.e., picking Al and then Bob is the same thing as picking Bob and then Al). So, we have to divide by 2!.
So we get: 12 * ((9*8)/2!) = 432.
Hope this helps!
Jim S. | GMAT Instructor | Veritas Prep
