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A simple combinatorics problem.
Word 'SUCCESS'. Number of ways such that no two Ss and Cs are together.
 one year ago
 one year ago
A simple combinatorics problem. Word 'SUCCESS'. Number of ways such that no two Ss and Cs are together.
 one year ago
 one year ago

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Ishaan94Best ResponseYou've already chosen the best response.3
@eliassaab @blockcolder
 one year ago

THE_PROPHETBest ResponseYou've already chosen the best response.0
my guess is 360....idk if i solved it rite....anyone??
 one year ago

THE_PROPHETBest ResponseYou've already chosen the best response.0
well its wrong sry!!
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.3
Stop guessing. Take your time and work on the problem.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.3
And for your information, this problem is not some challenge. I don't know how to do it. So you will have to show your work.
 one year ago

THE_PROPHETBest ResponseYou've already chosen the best response.0
dw:1340534838304:dw can be interchange so that no S are together in only 2 ways........ dw:1340534915656:dw 6 spaces can be filled by 2 Es in ....6c3/2 ways...... dw:1340534979616:dw.. so final answer is 15
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.3
No the answer is not 15.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
The complement of "no 2 S's and C's are together" is "2 S's or C's are together." 2 S's or C's together = 2 S's together + 2 C's together  2 S's and 2 C's together 2 S's together = 5! by considering the 3 s's as 1 2 C's together = 6! by considering 2 c's as 1 2 S's and 2 C's together = 4! So it's 7!(5!+6!4!)=4224 ways.
 one year ago

KRAZZyBest ResponseYou've already chosen the best response.0
Number of Letters/Characters in the word "SUCCESS" = 7 {S, U, C, C, E, S, S} No. of Letters : of the first kind ⇒ No. of S's = 3 ⇒ a = 3 of the second kind ⇒ No. of C's = 2 ⇒ b = 2 which are all different = 2 {U, E} ⇒ x = 2 = n=7!/3! × 2! = 7 × 6 × 5 × 4 × 3 × 2!/3 × 2 × 1 × 2! = 7 × 6 × 5 × 4 × 3 = 2,520
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.3
blockcoder you are including too many repetitive arrangements
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.0
answer cannot be more than 420
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.3
_S_S_S_ 4 spaces, 4 choices {C, C, E, U}. 4!/2! = 4*3 = 12. S_S_S__ > 12 __S_S_S > 12 So, 12 + 12 + 12 = 36. Haha Got it. Thanks all.
 one year ago

THE_PROPHETBest ResponseYou've already chosen the best response.0
yeah i got it thanks!!
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
What about S__S_S_ , S_S__S_?
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.3
I think, I included two repetitions
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.3
Ohh CCS_S_S and S_S_SCC isn't possible. It still comes out to be 36. Thanks Callisto.
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
If.... you have time.... and if ..... you don't mind, can you explain??
 one year ago
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