The number of 4 digit even numbers that can b | vedclass

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(C) To form a $4$ digit even number,the units place must be filled by $0, 2, 4,$ or $6$.Case $1$: Units place is $0$. The remaining $3$ places can be

Vedclass · Accepted Answer

$420$

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(C) To form a $4$ digit even number,the units place must be filled by $0, 2, 4,$ or $6$.Case $1$: Units place is $0$. The remaining $3$ places can be filled by the remaining $6$ digits $(1, 2, 3, 4, 5, 6)$ in $^6P_3 = 6 	imes 5 	imes 4 = 120$ ways.Case $2$: Units place is $2, 4,$ or $6$ ($3$ ways). The thousands place cannot be $0$ and cannot be the digit already used in the units place. Thus,there are $7 - 2 = 5$ choices for the thousands place. The remaining $2$ places can be filled by the remaining $5$ digits in $^5P_2 = 5 	imes 4 = 20$ ways. So,$3 	imes 5 	imes 20 = 300$ ways.Total number of ways $= 120 + 300 = 420$.CaseCalculationUnits place is $0$$1 	imes 6 	imes 5 	imes 4 = 120$Units place is ${2, 4, 6}$$3 	imes 5 	imes 5 	imes 4 = 300$Total$120 + 300 = 420$

The number of $4$ digit even numbers that can be formed using $0, 1, 2, 3, 4, 5, 6$ without repetition is

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