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22 Dec 2010, 08:22
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
71% (01:41) correct
29%
(01:45)
wrong
based on 190
sessions
History
Date
Time
Result
Not Attempted Yet
The botanic garden is planting bulbs for the spring flower show. The head gardener is trying to decide how to place three Tulips bulbs, three Daffodil bulbs, and three Crocus bulbs along the front row. How many different arrangements can he make of these bulbs in the front row?
(A) 27
(B) 108
(C) 216
(D) 512
(E) 1,680
(A) 27
(B) 108
(C) 216
(D) 512
(E) 1,680
22 Dec 2010, 09:10
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rxs0005
# of ways to arrange 9 flowers out of which 3 T's are identical, 3 D's are identical and 3 C's are identical is 9!/(3!3!3!)=1,680.
Answer: E.
Check this for more: counting-ps-106762.html#p840035
Hope it helps.
General Discussion
22 Dec 2010, 09:09
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total number of bulbs is 9, hence possible arrangements is 9!
But all the 3 bulbs are 3 each, hence
\(i.e., 9!/3!3!3! = 1680 ways\)
But all the 3 bulbs are 3 each, hence
\(i.e., 9!/3!3!3! = 1680 ways\)
