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# The botanic garden is planting bulbs for the spring flower

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Director
Joined: 07 Jun 2004
Posts: 614
Location: PA
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Kudos [?]: 315 [0], given: 22

The botanic garden is planting bulbs for the spring flower [#permalink]  22 Dec 2010, 07:22
00:00

Difficulty:

45% (medium)

Question Stats:

69% (02:07) correct 31% (02:09) wrong based on 16 sessions
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
[Reveal] Spoiler: OA

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Senior Manager
Status: swimming against the current
Joined: 24 Jul 2009
Posts: 252
Location: Chennai, India
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Kudos [?]: 49 [0], given: 30

Re: Combibations PS [#permalink]  22 Dec 2010, 08:09
total number of bulbs is 9, hence possible arrangements is 9!

But all the 3 bulbs are 3 each, hence

$$i.e., 9!/3!3!3! = 1680 ways$$
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Joined: 02 Sep 2009
Posts: 28227
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Kudos [?]: 44964 [0], given: 6637

Re: Combibations PS [#permalink]  22 Dec 2010, 08:10
Expert's post
rxs0005 wrote:
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

# 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.

Check this for more: counting-ps-106762.html#p840035

Hope it helps.
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Re: Combibations PS   [#permalink] 22 Dec 2010, 08:10
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