The total number of words (with or without me | vedclass

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(C) The word $'DISTRIBUTION'$ contains $12$ letters: $D, I, S, T, R, I, B, U, T, I, O, N$. Counting the frequency of each letter: $I: 3, T: 2, D: 1, S

Vedclass · Accepted Answer

$3734$

Vedclass · Answer

(C) The word $'DISTRIBUTION'$ contains $12$ letters: $D, I, S, T, R, I, B, U, T, I, O, N$. Counting the frequency of each letter: $I: 3, T: 2, D: 1, S: 1, R: 1, B: 1, U: 1, O: 1, N: 1$. There are $9$ distinct letters: ${D, I, S, T, R, B, U, O, N}$. We need to form words of length $4$. Case $1$: All $4$ letters are distinct. Number of ways $= {}^{9}C_{4} 	imes 4! = 126 	imes 24 = 3024$. Case $2$: $2$ letters are same and $2$ are distinct. Subcase $2a$: $I$ is repeated ($2$ $I$'s) and $2$ others from $8$ distinct letters. Number of ways $= {}^{8}C_{2} 	imes \frac{4!}{2!} = 28 	imes 12 = 336$. Subcase $2b$: $T$ is repeated ($2$ $T$'s) and $2$ others from $8$ distinct letters. Number of ways $= {}^{8}C_{2} 	imes \frac{4!}{2!} = 28 	imes 12 = 336$. Case $3$: $2$ pairs of same letters. Only $I$ and $T$ can form pairs. Number of ways $= \frac{4!}{2!2!} = 6$. Case $4$: $3$ letters are same and $1$ is distinct. Only $I$ can be selected $3$ times. Number of ways $= {}^{8}C_{1} 	imes \frac{4!}{3!} = 8 	imes 4 = 32$. Total number of words $= 3024 + 336 + 336 + 6 + 32 = 3734$.

The total number of words (with or without meaning) that can be formed out of the letters of the word $'DISTRIBUTION'$ taken four at a time,is equal to...............

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