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(B) Given digits are $0, 2, 4, 5$. Total number of four-digit numbers that can be formed without repetition: The first digit cannot be $0$,so there ar

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Given digits are $0, 2, 4, 5$. Total number of four-digit numbers that can be formed without repetition: The first digit cannot be $0$,so there are $3$ choices $(2, 4, 5)$. The remaining $3$ positions can be filled by the remaining $3$ digits in $3 	imes 2 	imes 1 = 6$ ways. Total numbers $= 3 	imes 3 	imes 2 	imes 1 = 18$. $A$ number is divisible by $5$ if it ends with $0$ or $5$. Case $1$: Number ends with $0$. The last digit is fixed as $0$. The remaining $3$ positions can be filled by the remaining $3$ digits $(2, 4, 5)$ in $3 	imes 2 	imes 1 = 6$ ways. Case $2$: Number ends with $5$. The last digit is fixed as $5$. The first digit cannot be $0$ or $5$,so there are $2$ choices $(2, 4)$. The remaining $2$ positions can be filled by the remaining $2$ digits in $2 	imes 1 = 2$ ways. Total numbers ending in $5 = 2 	imes 2 	imes 1 = 4$. Total numbers divisible by $5 = 6 + 4 = 10$. Total numbers not divisible by $5 = 18 - 10 = 8$.

The number of four-digit numbers formed by using the digits $0, 2, 4, 5$ (without repetition) which are not divisible by $5$ is:

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