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A simple combinatorics problem. Word 'SUCCESS'. Number of ways such that no two Ss and Cs are together.

Mathematics
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my guess is 360....idk if i solved it rite....anyone??
360!?! :o How?

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Other answers:

well its wrong sry!!
Of course it is.
396 ?
Noway Dude
factorial of 6??
Nope
Stop guessing. Take your time and work on the problem.
And for your information, this problem is not some challenge. I don't know how to do it. So you will have to show your work.
|dw:1340534838304:dw| can be interchange so that no S are together in only 2 ways........ |dw:1340534915656:dw| 6 spaces can be filled by 2 Es in ....6c3/2 ways...... |dw:1340534979616:dw|.. so final answer is 15
No the answer is not 15.
The complement of "no 2 S's and C's are together" is "2 S's or C's are together." 2 S's or C's together = 2 S's together + 2 C's together - 2 S's and 2 C's together 2 S's together = 5! by considering the 3 s's as 1 2 C's together = 6! by considering 2 c's as 1 2 S's and 2 C's together = 4! So it's 7!-(5!+6!-4!)=4224 ways.
Number of Letters/Characters in the word "SUCCESS" = 7 {S, U, C, C, E, S, S} No. of Letters : of the first kind ⇒ No. of S's = 3 ⇒ a = 3 of the second kind ⇒ No. of C's = 2 ⇒ b = 2 which are all different = 2 {U, E} ⇒ x = 2 = n=7!/3! × 2! = 7 × 6 × 5 × 4 × 3 × 2!/3 × 2 × 1 × 2! = 7 × 6 × 5 × 4 × 3 = 2,520
blockcoder you are including too many repetitive arrangements
answer cannot be more than 420
_S_S_S_ 4 spaces, 4 choices {C, C, E, U}. 4!/2! = 4*3 = 12. S_S_S__ -> 12 __S_S_S -> 12 So, 12 + 12 + 12 = 36. Haha Got it. Thanks all.
yeah i got it thanks!!
What about S__S_S_ , S_S__S_?
:(
I think, I included two repetitions
Ohh CCS_S_S and S_S_SCC isn't possible. It still comes out to be 36. Thanks Callisto.
If.... you have time.... and if ..... you don't mind, can you explain??

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