Number of ways the word 'SUCCESS' can be arranged, such that no two S's and C's are together.
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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.
oh i get it sorry
:( Byron left me an complicated answer. Integral and stuff.
siddhan can you post the solution as well? or, a hint.
Do you want the solution or hint?
Actually since you're not sure, I think I should post my solution.
I may be wrong.
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?
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