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(B) number is divisible by $55$ if it is divisible by both $5$ and $11$. For a number to be divisible by $5$,the last digit must be $0$ or $5$. Since

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$6$

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(B) number is divisible by $55$ if it is divisible by both $5$ and $11$. For a number to be divisible by $5$,the last digit must be $0$ or $5$. Since the available digits are ${2, 3, 4, 5, 6}$,the last digit must be $5$. Case $1$: $4$-digit numbers of the form $abc5$. For divisibility by $11$,the alternating sum of digits $(a+5) - (b+c)$ must be a multiple of $11$. Since $a, b, c \in {2, 3, 4, 6}$ and are distinct,the sum $a+b+c+5$ is at most $6+4+3+5 = 18$ and at least $2+3+4+5 = 14$. The possible values for $(a+5) - (b+c)$ are $0$ or $11$. If $(a+5) - (b+c) = 0$,then $a+5 = b+c$. Possible sets ${b, c}$ from ${2, 3, 4, 6}$ such that $b+c = a+5$: - If $a=2$,$b+c=7 \Rightarrow {3, 4}$. - If $a=3$,$b+c=8 \Rightarrow {2, 6}$. - If $a=4$,$b+c=9 \Rightarrow {3, 6}$. - If $a=6$,$b+c=11 \Rightarrow$ No pair. This gives $2$ permutations for each set: $(2, 3, 4, 5), (2, 4, 3, 5), (3, 2, 6, 5), (3, 6, 2, 5), (4, 3, 6, 5), (4, 6, 3, 5)$. Total $6$ numbers. Case $2$: $5$-digit numbers. The smallest $5$-digit number using these digits is $23456$,which is greater than $23421$. Thus,no $5$-digit number satisfies the condition. Therefore,the total count is $6$.

The number of natural numbers lying between $1012$ and $23421$ that can be formed using the digits $2, 3, 4, 5, 6$ (repetition of digits is not allowed) and divisible by $55$ is $....$

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