anonymous
  • anonymous
Number of ways the word 'SUCCESS' can be arranged, such that no two S's and C's are together. @Zarkon
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
And @siddhantsharan
anonymous
  • anonymous
\[7!/2!3!\]
anonymous
  • anonymous
Hold on. No. 120 - 24 = 96?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.