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Ishaan94
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A simple combinatorics problem.
Word 'SUCCESS'. Number of ways such that no two Ss and Cs are together.
 2 years ago
 2 years ago
Ishaan94 Group Title
A simple combinatorics problem. Word 'SUCCESS'. Number of ways such that no two Ss and Cs are together.
 2 years ago
 2 years ago

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Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
@eliassaab @blockcolder
 2 years ago

THE_PROPHET Group TitleBest ResponseYou've already chosen the best response.0
my guess is 360....idk if i solved it rite....anyone??
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
360!?! :o How?
 2 years ago

THE_PROPHET Group TitleBest ResponseYou've already chosen the best response.0
well its wrong sry!!
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
Of course it is.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
Noway Dude
 2 years ago

THE_PROPHET Group TitleBest ResponseYou've already chosen the best response.0
factorial of 6??
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
Stop guessing. Take your time and work on the problem.
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

THE_PROPHET Group TitleBest 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
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
No the answer is not 15.
 2 years ago

blockcolder Group TitleBest 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.
 2 years ago

KRAZZy Group TitleBest 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
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
blockcoder you are including too many repetitive arrangements
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
answer cannot be more than 420
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

THE_PROPHET Group TitleBest ResponseYou've already chosen the best response.0
yeah i got it thanks!!
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
What about S__S_S_ , S_S__S_?
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.3
I think, I included two repetitions
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
If.... you have time.... and if ..... you don't mind, can you explain??
 2 years ago
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