MHB How many words can be formed from the word SUCCESS without any repeated letters?

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The discussion focuses on calculating the number of distinct arrangements of the word "SUCCESS" without having any two 'C's or two 'S's adjacent. Initially, the total arrangements without restrictions are determined to be 420. The number of arrangements where two 'C's are together is calculated as 120, while the arrangements where no two 'S's are together is also found to be 120. By applying combinatorial methods, the final count of arrangements where no two 'C's and no two 'S's are together is concluded to be 96. The calculations demonstrate the application of permutations and combinations in solving the problem.
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how many number of words are formed from word $\bf{SUCCESS}$ such that no two $\bf{C}$ and no two $\bf{S}$ are together

My Trial:: First we will count Total no. of arrangement of words without restriction.

which is $\displaystyle = \frac{7!}{3!\times 2!} = 420$

Now Total no. of arrangement in which two $\bf{C}$ are together

which is $\displaystyle = \frac{6!}{3!} = 120$

Now Total no. of words in which no two $\bf{S}$ are togrther, is $\displaystyle = \binom{5}{3}\times \frac{4!}{2!} = 120$

Now I did understand How can i Calculate after that

So Help please

Thanks
 
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jacks said:
Now Total no. of words in which no two $\bf{S}$ are togrther, is $\displaystyle = \binom{5}{3}\times \frac{4!}{2!} = 120$
Subtract from that the number of words where no two S's are together, but both C's are together.
 
Thanks Evgeny.Makarov, Using your Hint:

Total no. of words in which two $\bf{S}$ are not together and two $\bf{C}$ are together

$ \displaystyle = \binom{4}{3}\times 3! = 24$

Now Total no. of words in which no two $\bf{C}$ and no two $\bf{S}$ are together , is $ = 120-24 = 96$

Got it.

Thanks
 
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