6 different letters of an alphabet are given. | vedclass

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Question

(A) Total words of $4$ letters with at least one repetition and no two same letters together can be formed in two cases:Case $1$: Two letters are same

Vedclass · Accepted Answer

$390$

Vedclass · Answer

(A) Total words of $4$ letters with at least one repetition and no two same letters together can be formed in two cases:Case $1$: Two letters are same and two are different (e.g.,$AABC$ type).Number of ways to select $1$ letter to be repeated: $^6C_1 = 6$.Number of ways to select $2$ other letters from remaining $5$: $^5C_2 = 10$.Number of ways to arrange $AABC$ such that no two same letters are together: $\frac{4!}{2!} 	imes 2 = 12 	imes 2 = 24$ (or by gap method: $3! 	imes ^3C_2 = 6 	imes 3 = 18$ is incorrect,correct is $12$ arrangements for $AABC$ where $AA$ are not together: Total $12$,$AA$ together $6$,so $12-6=6$. Wait,the arrangement of $AABC$ such that $AA$ are not together is $12$. Total arrangements of $AABC$ is $\frac{4!}{2!} = 12$. Arrangements with $AA$ together is $3! = 6$. So $12-6=6$. Total for Case $1$ is $6 	imes 10 	imes 6 = 360$.Case $2$: Two pairs of same letters (e.g.,$AABB$ type).Number of ways to select $2$ letters from $6$: $^6C_2 = 15$.Number of ways to arrange $AABB$ such that no two same letters are together: $2$ ways $(ABAB, BABA)$.Total for Case $2$ is $15 	imes 2 = 30$.Total number of words = $360 + 30 = 390$.

$6$ different letters of an alphabet are given. 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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