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[#39] 2 Min. Challenge: Counting Methods

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CEO
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[#39] 2 Min. Challenge: Counting Methods [#permalink]  27 Mar 2004, 19:45
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1. Time yourself
2. Solve as fast as you can

16 men are divided equally into 4 groups, how many such groupings can be made?

a. 16C4 *12C4 * 8C4 *4C4
b. 1/4! * 16C4* 12C4*8C4 *4C4
c. 1/4! * 16P4 * 12P4 *8P4 *4P4
d. 1/24 * 16P4 * 12P4*8P4 * 4P4
e. 16/ 4!
Senior Manager
Joined: 02 Mar 2004
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well, the groups are indistinguishable. So, the answer is B.
Director
Joined: 13 Nov 2003
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First group-16C4, second group-12C4, third group-8C4, last group-4C4. Rewritten-16!/4!4!4!4! or i believe A) is the answer
Senior Manager
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I thik it's A...reasoning with BG!!
CEO
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Kudos [?]: 701 [0], given: 781

Re: [#39] 2 Min. Challenge: Counting Methods [#permalink]  29 Mar 2004, 02:31
praetorian123 wrote:
1. Time yourself
2. Solve as fast as you can

16 men are divided equally into 4 groups, how many such groupings can be made?

a. 16C4 *12C4 * 8C4 *4C4
b. 1/4! * 16C4* 12C4*8C4 *4C4
c. 1/4! * 16P4 * 12P4 *8P4 *4P4
d. 1/24 * 16P4 * 12P4*8P4 * 4P4
e. 16/ 4!

halle. is correct.

had the groups been different from each other, A would have been the correct answer. but since the groups are indistinguishable , we have to divide A by 4! ...which gives us B.
Director
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Posts: 793
Location: BULGARIA
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Hallo Pret, can you please explain what "indistinguishable" means, because i can not understand this part. Thanks.
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[#39] 2 Min. Challenge: Counting Methods

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