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Number of ways the word 'SUCCESS' can be arranged, such that no two S's and C's are together. @Zarkon

Mathematics
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\[7!/2!3!\]
Hold on. No. 120 - 24 = 96?

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Other answers:

I can't be sure of the answer. To convince me you would need to show working as well.
In my textbook 36 is the given answer. But I don't think it's right, now that siddhant and another guy I knew answered 96 instead.
no. of letters= 7 no. of s = 3 no. of c = 2 therfore no. of combination= 7!/2!3!
Chumku read the question again.
know*
oh i get it sorry
lol :( Byron left me an complicated answer. Integral and stuff.
a*
siddhan can you post the solution as well? or, a hint.
siddhant*
Yeahhh. Sorry. Do you want the solution or hint?
Actually since you're not sure, I think I should post my solution. I may be wrong. So, No. of cases with no two C's together and no two S's together = No. of cases with no two S's together ( C's can me arranged anyhow) - No. of cases with no two S's together ( C's being together now)
Does that seem correct so far? @Ishaan94
It does
|dw:1340389706521:dw| Writing the letters with spaces between them. And only putting S's in the spaces so as to ensure their seperation. Total ways of doing this ( S's together C's anyhow) : 5C2 * 4!/2!
And it also proves how simple this problem could be
And how stupid I think at times -sigh-
Yeah. It's pretty simple now.
Thank you
Anytime :)

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