- #1
juantheron
- 247
- 1
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
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