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# How many ways are possible to arrange A, B, C, C, and D with

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How many ways are possible to arrange A, B, C, C, and D with [#permalink]  20 Apr 2008, 11:48
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How many ways are possible to arrange A, B, C, C, and D with two "C" being separated by at least one letter?
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Re: PS: Possible ways [#permalink]  20 Apr 2008, 11:56
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I get 36

Perms:
CACBD - 6 possibilities
CABCD - 6
CABDC-6
ACBCD-6
ACBDC-6
ABCDC-6

6*6 = 36

However if Cs are distinct (which is probably not what the question implies), the answer would be 72.
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Re: PS: Possible ways [#permalink]  20 Apr 2008, 11:57
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C _ C _ _
C _ _ C _
C _ _ _ C
_ C _ C _
_ C _ _ C
_ _ C _ C

In each of these, we have 3 blanks, the contents of which are completely up for variation (thus it is a factorial).

3! = 3*2*1 = 6 possibilities for each of the patterns above

6*6 = 36

Another note:
--Since the C's are the same, we don't need to account for switching the two C's in each case above.
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Re: PS: Possible ways [#permalink]  20 Apr 2008, 22:03
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I get 96 (5!-4!) , is that correct?

5! = number of ways to arrange A,B,C,C,D
4! = assuming CC are consecutive

so Number of ways such that C,C are atlease one space apart is 5!-4!=96
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Kudos [?]: 2117 [1] , given: 359

Re: PS: Possible ways [#permalink]  20 Apr 2008, 22:15
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Expert's post
rpmodi wrote:
I get 96 (5!-4!) , is that correct?

5! = number of ways to arrange A,B,C,C,D
4! = assuming CC are consecutive

so Number of ways such that C,C are atlease one space apart is 5!-4!=96

5! corresponds to 5 different things but here we have the same two things and 3 different. So, the correct total number of ways to arrange A,B,C,C,D will be 5!/2. ($$A,B,C_1,C_2,D$$ and $$A,B,C_2,C_1,D$$ is the same combination)

5!/2-4!=60-24=36
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Re: PS: Possible ways [#permalink]  20 Apr 2008, 22:16
sorry , my answer is 36 as well

it should be 5!/2 - 4! ( 5!/2 since we have a repetition of C )
Re: PS: Possible ways   [#permalink] 20 Apr 2008, 22:16
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