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(B) Since the required numbers are less than $1000$,they can be $1$-digit,$2$-digit,or $3$-digit numbers.$(i)$ For $1$-digit odd numbers: The unit's p

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$32$

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(B) Since the required numbers are less than $1000$,they can be $1$-digit,$2$-digit,or $3$-digit numbers.$(i)$ For $1$-digit odd numbers: The unit's place must be an odd digit. The available odd digits are $5$ and $7$. Thus,there are $2$ possible $1$-digit odd numbers.(ii) For $2$-digit odd numbers: The unit's place can be filled by $5$ or $7$ ($2$ ways). The ten's place can be filled by any digit except $0$,i.e.,$2, 5,$ or $7$ ($3$ ways). Total $2$-digit odd numbers $= 2 	imes 3 = 6$.(iii) For $3$-digit odd numbers: The unit's place can be filled by $5$ or $7$ ($2$ ways). The ten's place can be filled by any of the $4$ digits $(0, 2, 5, 7)$ ($4$ ways). The hundred's place can be filled by any digit except $0$,i.e.,$2, 5,$ or $7$ ($3$ ways). Total $3$-digit odd numbers $= 2 	imes 4 	imes 3 = 24$.Total number of odd numbers $= 2 + 6 + 24 = 32$.

How many odd numbers less than $1000$ can be formed using the digits $0, 2, 5, 7$? (repetition of digits is allowed).

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