Ishaan94
Number of ways the word 'SUCCESS' can be arranged, such that no two S's and C's are together.
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Ishaan94
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And @siddhantsharan
Chumku
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\[7!/2!3!\]
siddhantsharan
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Hold on. No. 120 - 24 = 96?
Ishaan94
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I can't be sure of the answer. To convince me you would need to show working as well.
Ishaan94
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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.
Chumku
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no. of letters= 7
no. of s = 3
no. of c = 2
therfore no. of combination= 7!/2!3!
Ishaan94
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Chumku read the question again.
Ishaan94
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know*
Chumku
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oh i get it sorry
Ishaan94
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lol
:( Byron left me an complicated answer. Integral and stuff.
Ishaan94
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a*
Ishaan94
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siddhan can you post the solution as well? or, a hint.
Ishaan94
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siddhant*
siddhantsharan
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Yeahhh.
Sorry.
Do you want the solution or hint?
siddhantsharan
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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)
siddhantsharan
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Does that seem correct so far?
@Ishaan94
Ishaan94
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It does
siddhantsharan
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|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!
Ishaan94
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And it also proves how simple this problem could be
Ishaan94
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And how stupid I think at times -sigh-
siddhantsharan
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Yeah. It's pretty simple now.
Ishaan94
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Thank you
siddhantsharan
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Anytime :)