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 $4$ distinct

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$240$

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 and $2$ distinct letters.Step $1$: Select the pair of identical letters. There are $2$ pairs ($S, S$ or $L, L$),so we can choose $1$ pair in $^2C_1 = 2$ ways.Step $2$: Select $2$ distinct letters from the remaining $4$ distinct letters $(Y, A, B, U)$. We can choose $2$ letters in $^4C_2 = 6$ ways.Step $3$: Arrange these $4$ 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 = $(	ext{Ways to select pair}) 	imes (	ext{Ways to select distinct letters}) 	imes (	ext{Arrangements}) = 2 	imes 6 	imes 12 = 144$.Wait,re-evaluating: The distinct letters available are $Y, A, B, U$ and the other pair. If we pick $S, S$,we have $L, Y, A, B, U$ to pick $2$ from. Total distinct letters available are $5$ $(L, Y, A, B, U)$.Correct calculation: Select $1$ pair from $2$ ($S, S$ or $L, L$) = $2$ ways. Select $2$ distinct letters from the remaining $5$ letters = $^5C_2 = 10$ ways.Total 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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