The number of words,with or without meaning,t | vedclass

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(D) The word $'SYLLABUS'$ contains $8$ letters: $S, S, L, L, Y, A, B, U$. There are $2$ pairs of identical letters ($S, S$ and $L, L$) and $4$ distinc

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(D) The word $'SYLLABUS'$ contains $8$ letters: $S, S, L, L, Y, A, B, U$. There are $2$ pairs of identical letters ($S, S$ and $L, L$) and $4$ distinct letters $(Y, A, B, U)$. We need to form a $4$-letter word with $2$ alike letters and $2$ distinct letters. Step $1$: Select the pair of identical letters. There are $2$ pairs available ($S$ or $L$),so we choose $1$ pair in $^2C_1 = 2$ ways. Step $2$: Select $2$ distinct letters from the remaining $4$ distinct letters $(Y, A, B, U)$ and the $1$ remaining pair (treating it as distinct letters). Actually,we have $5$ distinct types of letters available $(S, L, Y, A, B, U)$ after picking $1$ pair. Since we picked $1$ pair,we have $5$ remaining types to pick $2$ distinct letters from. So,$^5C_2 = 10$ ways. Step $3$: Arrange the $4$ letters (where $2$ are alike and $2$ are distinct). The number of arrangements is $\frac{4!}{2!} = \frac{24}{2} = 12$. Total number of words = $2 	imes 10 	imes 12 = 240$.

The number of words,with or without meaning,that can be formed by taking $4$ letters at a time from the letters of the word $'SYLLABUS'$ such that two letters are distinct and two letters are alike,is

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