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How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’

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Math Revolution GMAT Instructor
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How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’  [#permalink]

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New post 13 Feb 2018, 01:39
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

76% (01:04) correct 24% (01:35) wrong based on 109 sessions

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[GMAT math practice question]

How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ in a straight line are possible?

A. 60
B. 120
C. 180
D. 240
E. 420

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How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’  [#permalink]

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New post 13 Feb 2018, 01:58
MathRevolution wrote:
[GMAT math practice question]

How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ in a straight line are possible?

A. 60
B. 120
C. 180
D. 240
E. 420



There is a total of 7 alphabets, having 2 alphabets which repeat more than once - 2 'c' and 3 's'.

Therefore, the total number of arrangements possible is \(\frac{7!}{2!*3!} = \frac{7*6*5*4*3*2}{3*4} = 7*6*5*2\) = 420(Option E)
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Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
Posts: 7504
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GPA: 3.82
Re: How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’  [#permalink]

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New post 15 Feb 2018, 04:40
=>

The seven letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ include three s’s and two c’s.
The number of permutations is 7! / (3!*2!) = 420.

Therefore, the answer is E.

Answer: E
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Re: How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’  [#permalink]

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New post 11 Sep 2018, 09:59
1
MathRevolution wrote:
[GMAT math practice question]

How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ in a straight line are possible?

A. 60
B. 120
C. 180
D. 240
E. 420



Dear Moderator,
Please untag probability, Thank you.
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Re: How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’  [#permalink]

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New post 11 Sep 2018, 10:09
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Re: How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’  [#permalink]

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New post 12 Sep 2018, 18:30
MathRevolution wrote:
[GMAT math practice question]

How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ in a straight line are possible?

A. 60
B. 120
C. 180
D. 240
E. 420


We recognize that the 3 s’s and the 2 c’s are indistinguishable. Thus, the number of ways to arrange ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ in a straight line is:

7!/(3! x 2!) = (7 x 6 x 5 x 4)/2! = 7 x 6 x 5 x 2 = 420

Answer: E
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Re: How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’   [#permalink] 12 Sep 2018, 18:30
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