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(C) The word $SUCCESS$ contains the letters: $S, S, S, U, C, C, E$.Total letters: $7$ $(S:3, C:2, U:1, E:1)$.We need to arrange these such that the tw

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$24$

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(C) The word $SUCCESS$ contains the letters: $S, S, S, U, C, C, E$.Total letters: $7$ $(S:3, C:2, U:1, E:1)$.We need to arrange these such that the two $C$ are together and no two $S$ are together.Treat $(CC)$ as a single unit. The remaining letters are $U, E$.Arrange the units $(CC), U, E$ in $3!$ ways.These $3$ units create $4$ gaps: $\_ (CC) \_ U \_ E \_$.We have $3$ $S$ to place in these $4$ gaps such that no two $S$ are together. This can be done in $^4C_3$ ways.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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