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(A) Total words of $4$ letters with at least one letter repeated and no two same letters are together can be calculated by considering two cases for t

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$390$

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(A) Total words of $4$ letters with at least one letter repeated and no two same letters are together can be calculated by considering two cases for the selection of letters:Case $1$: Two letters are the same (one pair) and two letters are different.Number of ways to select the pair: $^6C_1 = 6$.Number of ways to select the other two distinct letters: $^5C_2 = 10$.Number of ways to arrange these $4$ letters $(A, A, B, C)$ such that no two same letters are together:Total arrangements of $A, A, B, C$ is $\frac{4!}{2!} = 12$.Arrangements where $A, A$ are together: Treat $(AA)$ as one unit,so we arrange $(AA), B, C$ in $3! = 6$ ways.Valid arrangements = $12 - 6 = 6$.Total for Case $1 = 6 	imes 10 	imes 6 = 360$.Case $2$: Two pairs of same letters (e.g.,$A, A, B, B$).Number of ways to select two pairs: $^6C_2 = 15$.Number of ways to arrange $A, A, B, B$ such that no two same letters are together:Total arrangements = $\frac{4!}{2!2!} = 6$.Arrangements where $AA$ are together: $3! = 6$.Arrangements where $BB$ are together: $3! = 6$.Arrangements where both $AA$ and $BB$ are together: $2! = 2$.Using inclusion-exclusion: $6 - (6 + 6) + 2 = -2$ (Wait,let's list them: $ABAB, ABBA, BAAB, BABA$ - there are $2$ valid arrangements).Total for Case $2 = 15 	imes 2 = 30$.Total = $360 + 30 = 390$.

Given $6$ different letters of an alphabet. Words with $4$ letters are formed from these given letters. The number of words which have at least one letter repeated and no two same letters are together,is:

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