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(C) The total number of ways to form a four-digit number using $7$ distinct digits is given by $P(7, 4) = 7 	imes 6 	imes 5 	imes 4 = 840$. For the

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$4/7$

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(C) The total number of ways to form a four-digit number using $7$ distinct digits is given by $P(7, 4) = 7 	imes 6 	imes 5 	imes 4 = 840$. For the number to be $> 4000$,the first digit (thousands place) must be $4, 5, 6,$ or $7$. There are $4$ choices for the first digit. After choosing the first digit,the remaining $3$ positions can be filled by the remaining $6$ digits in $P(6, 3) = 6 	imes 5 	imes 4 = 120$ ways. Thus,the number of favourable cases is $4 	imes 120 = 480$. Therefore,the required probability is $\frac{480}{840} = \frac{4}{7}$.

$A$ four-digit number is to be formed using the digits $1, 2, 3, 4, 5, 6, 7$ (no digit is repeated). What is the probability that the number formed is $> 4000$?

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