The number of different words that can be for | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(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:

Similar Questions

The remainder of $n^4-2n^3-n^2+2n-26$ when divided by $24$ is:

The number of $6$-digit numbers that can be formed using the digits $0, 1, 2, 5, 7,$ and $9$ which are divisible by $11$,where no digit is repeated,is:

The number of $9$-digit even natural numbers formed using only the digits $0$ and $1$,such that no two consecutive digits are $0$,is:

The number of ways in which four letters of the word '$MATHEMATICS$' can be arranged is given by

The sum of all $4$-digit numbers that can be formed using the digits $2, 3, 4, 5, 6$ without repetition is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module