Among the 4-digit numbers that can be formed | vedclass

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(A) number is divisible by $6$ if it is divisible by both $2$ and $3$.For a number to be divisible by $2$,the units digit must be $2, 4,$ or $6$.For a

Vedclass · Accepted Answer

$60$

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(A) number is divisible by $6$ if it is divisible by both $2$ and $3$.For a number to be divisible by $2$,the units digit must be $2, 4,$ or $6$.For a number to be divisible by $3$,the sum of its digits must be divisible by $3$.Let the $4$ digits be $d_1, d_2, d_3, d_4$ where $d_4 \in \{2, 4, 6\}$.Case $I$: Units digit is $2$. The sum of the remaining $3$ digits must be $3k - 2$. Possible sets of $3$ digits from $\{1, 3, 4, 5, 6\}$ are $\{1, 4, 6\}, \{1, 5, 6\}, \{3, 4, 5\}$. Each set can be arranged in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $II$: Units digit is $4$. The sum of the remaining $3$ digits must be $3k - 4$. Possible sets of $3$ digits from $\{1, 2, 3, 5, 6\}$ are $\{1, 2, 6\}, \{1, 5, 6\}, \{2, 3, 6\}$. Each set can be arranged in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $III$: Units digit is $6$. The sum of the remaining $3$ digits must be $3k - 6$. Possible sets of $3$ digits from $\{1, 2, 3, 4, 5\}$ are $\{1, 2, 3\}, \{1, 3, 5\}, \{2, 3, 4\}, \{3, 4, 5\}$. Each set can be arranged in $3! = 6$ ways. Total $= 4 	imes 6 = 24$.Total numbers $= 18 + 18 + 24 = 60$.

Among the $4$-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6$ without repeating any digit,the number of numbers which are divisible by $6$ is:

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