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(D) To find the number of $5$-digit odd numbers greater than $40,000$ with at least one digit repeated,we calculate: (Total $5$-digit odd numbers $> 4

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$2352$

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(D) To find the number of $5$-digit odd numbers greater than $40,000$ with at least one digit repeated,we calculate: (Total $5$-digit odd numbers $> 40,000$) - (Total $5$-digit odd numbers $> 40,000$ with no repetition).Step $1$: Total $5$-digit odd numbers $> 40,000$ (with repetition allowed).The first digit can be $4, 5, 6,$ or $7$ ($4$ choices).The last digit must be $3, 5,$ or $7$ ($3$ choices).The second,third,and fourth digits can be any of the $6$ digits ($6$ choices each).Total $= 4 	imes 6 	imes 6 	imes 6 	imes 3 = 2592$.Step $2$: Total $5$-digit odd numbers $> 40,000$ (no repetition).Case $1$: Last digit is $3$. First digit can be $4, 5, 6, 7$ ($4$ choices). Remaining $3$ positions filled by remaining $4$ digits: $4 	imes 4 	imes 3 	imes 2 = 96$.Case $2$: Last digit is $5$ or $7$ ($2$ choices). First digit can be $4, 6$ (if $5$ is used) or $4, 5, 6$ (if $7$ is used). If last digit is $5$,first digit can be $4, 6, 7$ ($3$ choices). Remaining $3$ positions: $3 	imes 4 	imes 3 	imes 2 = 72$.If last digit is $7$,first digit can be $4, 5, 6$ ($3$ choices). Remaining $3$ positions: $3 	imes 4 	imes 3 	imes 2 = 72$.Total without repetition $= 96 + 72 + 72 = 240$.Step $3$: Required numbers $= 2592 - 240 = 2352$.

The number of $5$-digit odd numbers greater than $40,000$ that can be formed by using the digits $3, 4, 5, 6, 7, 0$ such that at least one of its digits is repeated is:

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