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# [#39] 2 Min. Challenge: Counting Methods

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[#39] 2 Min. Challenge: Counting Methods [#permalink]

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27 Mar 2004, 20:45
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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!
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27 Mar 2004, 20:51
well, the groups are indistinguishable. So, the answer is B.
Director
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28 Mar 2004, 04:03
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
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28 Mar 2004, 05:55
I thik it's A...reasoning with BG!!
CEO
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Re: [#39] 2 Min. Challenge: Counting Methods [#permalink]

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29 Mar 2004, 03:31
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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29 Mar 2004, 03:39
Hallo Pret, can you please explain what "indistinguishable" means, because i can not understand this part. Thanks.
29 Mar 2004, 03:39
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