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There are $5$ one-digit natural numbers out of the given digits.

For getting $2−digit$ natural numbers we cannot put $0$ at the ten's place. So, this place can be filled by any of the given five nonzero digits in $5$ ways.

The unit's digit can be filled by any of the given six digits in $6$ ways.

So, the number of $2−digit$ natural numbers $=(5×6)=30$.

Similarly, to get a $3−digit$ number, we cannot put on at the hundred's place. So, this place can be filled in $5$ ways, each of the ten's and unit's places can be filled in $6$ ways.

$∴$ the number of $3−digit$ numbers $=(5×6×6)=180$.

Required number of numbers $=(5+30+180)=215$.

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