The number of different words that can be for | vedclass

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Question

(C) The word $SUCCESS$ contains $7$ letters: $S, S, S, C, C, U, E$. Step $1$: Treat the two $C$s as a single unit $(CC)$. The remaining letters are $S

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) The word $SUCCESS$ contains $7$ letters: $S, S, S, C, C, U, E$. Step $1$: Treat the two $C$s as a single unit $(CC)$. The remaining letters are $S, S, S, U, E$. Step $2$: Arrange the letters $S, S, S, U, E$ such that no two $S$ are together. First,arrange the non-$S$ letters: $(CC), U, E$. These $3$ items can be arranged in $3! = 6$ ways. Step $3$: These $3$ items create $4$ gaps: $\_ (CC) \_ U \_ E \_$. Step $4$: We need to place $3$ $S$s into these $4$ gaps so that no two $S$s are together. This can be done in $^4C_3$ ways. Step $5$: Total number of words $= 3! 	imes ^4C_3 = 6 	imes 4 = 24$.

The number of different words that can be formed from the letters of the word $SUCCESS$ in which the two $C$ are together but no two $S$ are together are:

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