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The number of arrangements that can be made out of the letters of the

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SUV0508
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The number of arrangements that can be made out of the letters of the [#permalink] The number of arrangements that can be made out of the letters of the   31 Jul 2022, 22:38
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The number of arrangements that can be made out of the letters of the word SUCCESS so that all the S's do not come together is:

(A) 60
(B) 120
(C) 360
(D) 420
(E) 450


P.S. I know how to solve it. However, I was reading about the gap method and am wondering if the problem could be solved using it
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C
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sanjitscorps18
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Re: The number of arrangements that can be made out of the letters of the [#permalink] The number of arrangements that can be made out of the letters of the   01 Sep 2022, 12:24
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All the combination of the word SUCCESS are 7!/(3!*2!) = 420

Considering all the Ss together, the number of combinations possible are 5!/2! = 60

Combinations where all the Ss are not together are 420 - 60 = 360

Option C

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Re: The number of arrangements that can be made out of the letters of the [#permalink] The number of arrangements that can be made out of the letters of the   01 Sep 2022, 19:43
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SUV0508 wrote:
The number of arrangements that can be made out of the letters of the word SUCCESS so that all the S's do not come together is:

(A) 60
(B) 120
(C) 360
(D) 420
(E) 450


P.S. I know how to solve it. However, I was reading about the gap method and am wondering if the problem could be solved using it


Number of arrangements that can be made so that all the S's DO NOT come together = Total normal arrangements - Number of arrangements in which all S's DO COME TOGETHER

Normal arrangements = \(\frac{7!}{2!3!}\)
Number of arrangements in which all S's ARE TOGETHER - (SSS)UCCE; Considering the 3 S's in the bracket as a single unit so arrangements = \(\frac{5!}{2!}\)

Required arrangements = \(\frac{7!}{2!3!} - \frac{5!}{2!} = 420 - 60 = 360\)

Answer - C
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aj1410
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Re: The number of arrangements that can be made out of the letters of the [#permalink] The number of arrangements that can be made out of the letters of the   07 Sep 2022, 00:03
Hi, in this case, why we have not taken 3! for three S' when these 3 S are together and why only 2! for C only?

Pls help.

Thanks :)
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Re: The number of arrangements that can be made out of the letters of the [#permalink] The number of arrangements that can be made out of the letters of the   28 Sep 2022, 14:37
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aj1410 wrote:
Hi, in this case, why we have not taken 3! for three S' when these 3 S are together and why only 2! for C only?

Pls help.

Thanks :)


The \(2!\) is to account for doublecounting of the two Cs in the letter pool. The three Ss are looked at as one unit or letter and are represented in the \(5!\) in the numerator.
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Re: The number of arrangements that can be made out of the letters of the [#permalink] The number of arrangements that can be made out of the letters of the   03 Feb 2024, 14:48
If we pull together all the S's for our null; i.e., SSS, why do we not "unlink" them at the end by multiplying the 60 by 3! ?.

total permutations = 420
permutations w/ S,S,S = 60 * 3! = 360. (??)
420-360 = 60.

Could someone explain? I thought when you linked together items in a permutation, you had to "unlink" them by multiplying by x! afterward - where (x) = the number of linked items.

Does this not apply here because the S's are indistinguishable items?
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The number of arrangements that can be made out of the letters of the [#permalink] The number of arrangements that can be made out of the letters of the   04 Feb 2024, 07:01
griffith7 wrote:
If we pull together all the S's for our null; i.e., SSS, why do we not "unlink" them at the end by multiplying the 60 by 3! ?.

total permutations = 420
permutations w/ S,S,S = 60 * 3! = 360. (??)
420-360 = 60.

Could someone explain? I thought when you linked together items in a permutation, you had to "unlink" them by multiplying by x! afterward - where (x) = the number of linked items.

Does this not apply here because the S's are indistinguishable items?


The total possible permutations of the 7 letters is:

7!/3!2! = 420

The S's already have their 3! permutations embedded in the numerator factored out by the 3! in the denominator so it is unnecessary to gross up the 3 S's by 3!....they are already on the same footing.

In other words, the 3 S's in the 420 are not permuted among themselves so the number to be subtracted from 420 should not include a 3! permutation of 3 S's.

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