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(B) number is divisible by $25$ if its last two digits are $25, 50,$ or $75$. Given the set of digits $\{1, 2, 3, 4, 5, 6, 7\}$,the possible two-digit

Vedclass · Accepted Answer

$40$

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(B) number is divisible by $25$ if its last two digits are $25, 50,$ or $75$. Given the set of digits $\{1, 2, 3, 4, 5, 6, 7\}$,the possible two-digit endings divisible by $25$ are $25$ and $75$ (since $50$ is not possible as $0$ is not in the set). Case $1$: The number ends in $25$. The last two digits are fixed as $2$ and $5$. The remaining $2$ positions can be filled by the remaining $5$ digits $\{1, 3, 4, 6, 7\}$ in $P(5, 2) = 5 	imes 4 = 20$ ways. Case $2$: The number ends in $75$. The last two digits are fixed as $7$ and $5$. The remaining $2$ positions can be filled by the remaining $5$ digits $\{1, 2, 3, 4, 6\}$ in $P(5, 2) = 5 	imes 4 = 20$ ways. Total numbers divisible by $25 = 20 + 20 = 40$.

If four-digit numbers are formed by using the digits $1, 2, 3, 4, 5, 6, 7$ without repetition,then how many of these numbers are exactly divisible by $25$?

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