The number of three-digit even numbers,formed | vedclass

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(B) To form a three-digit even number,the unit place must be occupied by an even digit $(0, 4, 6)$. Since repetition is not allowed,we consider two ca

Vedclass · Accepted Answer

$52$

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(B) To form a three-digit even number,the unit place must be occupied by an even digit $(0, 4, 6)$. Since repetition is not allowed,we consider two cases:Case $(i)$: When the unit place is $0$.There is $1$ way to fill the unit place. The remaining two places (hundreds and tens) can be filled by the remaining $5$ digits $(1, 3, 4, 6, 7)$ in $5 	imes 4 = 20$ ways.Case $(ii)$: When the unit place is $4$ or $6$.There are $2$ ways to fill the unit place. The hundreds place cannot be $0$ and cannot be the digit already used in the unit place,so there are $6 - 2 = 4$ choices for the hundreds place. The tens place can then be filled by any of the remaining $4$ digits (including $0$),giving $4 	imes 4 = 16$ ways for each choice of the unit digit. Total ways $= 2 	imes 16 = 32$ ways.Total number of even numbers $= 20 + 32 = 52$.

The number of three-digit even numbers,formed by the digits $0, 1, 3, 4, 6, 7$ if the repetition of digits is not allowed,is .... .

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