The number of six-digit natural numbers that | vedclass

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(D) We need to form a six-digit number using the digits $\{0, 2, 3, 4, 5, 6, 7, 8\}$.There are $8$ available digits in total.For a six-digit number,th

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$7 	imes 2^{15}$

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(D) We need to form a six-digit number using the digits $\{0, 2, 3, 4, 5, 6, 7, 8\}$.There are $8$ available digits in total.For a six-digit number,the first digit (at the hundred-thousands place) cannot be $0$. Thus,there are $7$ choices for the first digit $(\{2, 3, 4, 5, 6, 7, 8\})$.Since repetition of digits is allowed,each of the remaining $5$ positions can be filled by any of the $8$ available digits.Therefore,the total number of six-digit numbers is:$7 	imes 8 	imes 8 	imes 8 	imes 8 	imes 8 = 7 	imes 8^5$Since $8 = 2^3$,we have $8^5 = (2^3)^5 = 2^{15}$.Thus,the total number of such six-digit numbers is $7 	imes 2^{15}$.

The number of six-digit natural numbers that can be formed using the digits $2, 3, 4, 0, 5, 6, 7, 8$ (repetition of digits is allowed) is:

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