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arrangement of word SUCCESS

jacks

Well-known member
Apr 5, 2012
226
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
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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.
 

jacks

Well-known member
Apr 5, 2012
226
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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