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yb
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I get 144 for this.. 12 experienced + 9 others. If we are to 21 Dec 2005, 15:16
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I get 144 for this..

12 experienced + 9 others.

If we are to have exactly 1 experience plus 2 others,

(12*9*8)/3! = 144 different groups are possible. Can someone explain where I am going wrong? Thanks



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SunShine
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I get 144 for this.. 12 experienced + 9 others. If we are to 21 Dec 2005, 16:31
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432.

1 experience to chose form 12, 12C1 = 12
2 Others to chose from 9, 9C2 = 36

together, 12*36 = 432
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TeHCM
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I get 144 for this.. 12 experienced + 9 others. If we are to 21 Dec 2005, 16:43
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There are twelve ways that you could pick an experienced person, and there are 9 that are inexperienced.

12 * 9C2 = 432
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HIMALAYA
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I get 144 for this.. 12 experienced + 9 others. If we are to 21 Dec 2005, 22:12
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agreed. all are good ones.
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vivek123
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I get 144 for this.. 12 experienced + 9 others. If we are to 21 Dec 2005, 22:30
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Number of ways of selecting an experience member = 12.
Number os ways of selecting remaining 2 members = 9C2

Total = 12 * 9C2 = 432.
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ywilfred
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I get 144 for this.. 12 experienced + 9 others. If we are to 22 Dec 2005, 00:58
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21 astronauts -- 12 experienced, 9 inexperienced

3 person crew so that exactly 1 is experienced

# of ways to pick 2 inexperienced = 9C2 = 36
This 36 can be teamed up with 12 different experienced astronauts.

total # of ways = 36*12 = 432



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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