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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hold on. No. 120  24 = 96?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can't be sure of the answer. To convince me you would need to show working as well.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no. of letters= 7 no. of s = 3 no. of c = 2 therfore no. of combination= 7!/2!3!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Chumku read the question again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol :( Byron left me an complicated answer. Integral and stuff.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0siddhan can you post the solution as well? or, a hint.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeahhh. Sorry. Do you want the solution or hint?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does that seem correct so far? @Ishaan94

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And it also proves how simple this problem could be

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And how stupid I think at times sigh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah. It's pretty simple now.
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