The number of natural numbers less than 7,000 | vedclass

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(B) We need to form natural numbers less than $7,000$ using the digits ${0, 1, 3, 7, 9}$ with repetition allowed.Case $1$: $1$-digit,$2$-digit,or $3$-

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$374$

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(B) We need to form natural numbers less than $7,000$ using the digits ${0, 1, 3, 7, 9}$ with repetition allowed.Case $1$: $1$-digit,$2$-digit,or $3$-digit numbers.For a $1$-digit number,there are $4$ choices $(1, 3, 7, 9)$ since $0$ cannot be the first digit.For a $2$-digit number,the first digit has $4$ choices $(1, 3, 7, 9)$ and the second has $5$ choices $(0, 1, 3, 7, 9)$,so $4 	imes 5 = 20$ numbers.For a $3$-digit number,the first digit has $4$ choices and the next two have $5$ choices each,so $4 	imes 5 	imes 5 = 100$ numbers.Total for $1, 2, 3$ digits $= 4 + 20 + 100 = 124$.Case $2$: $4$-digit numbers less than $7,000$.The first digit can be $1$ or $3$ (since it must be less than $7$).If the first digit is $1$ or $3$ ($2$ choices),the remaining $3$ positions can each be filled in $5$ ways.Number of such $4$-digit numbers $= 2 	imes 5 	imes 5 	imes 5 = 250$.Total numbers $= 124 + 250 = 374$.

The number of natural numbers less than $7,000$ which can be formed by using the digits $0, 1, 3, 7, 9$ (repetition of digits allowed) is equal to

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