The number of integers greater than 3000 that | vedclass

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(B) To form integers greater than $3000$ using digits $\{0, 1, 2, 3, 4, 5\}$ without repetition,we consider numbers with $4, 5,$ and $6$ digits.$1$. $

Vedclass · Accepted Answer

$1380$

Vedclass · Answer

(B) To form integers greater than $3000$ using digits $\{0, 1, 2, 3, 4, 5\}$ without repetition,we consider numbers with $4, 5,$ and $6$ digits.$1$. $4$-digit numbers: The first digit must be $3, 4,$ or $5$ ($3$ choices). The remaining $3$ positions can be filled by the remaining $5$ digits in $P(5, 3) = 5 	imes 4 	imes 3 = 60$ ways. Total = $3 	imes 60 = 180$. However,if the first digit is $3$,the number is $3xyz$. If $x=0$,we have $30yz$,which is $> 3000$. If $x > 0$,it is also $> 3000$. So all $180$ are valid.$2$. $5$-digit numbers: The first digit cannot be $0$ ($5$ choices: $1, 2, 3, 4, 5$). The remaining $4$ positions can be filled by the remaining $5$ digits in $P(5, 4) = 5 	imes 4 	imes 3 	imes 2 = 120$ ways. Total = $5 	imes 120 = 600$.$3$. $6$-digit numbers: The first digit cannot be $0$ ($5$ choices). The remaining $5$ positions can be filled by the remaining $5$ digits in $P(5, 5) = 5! = 120$ ways. Total = $5 	imes 120 = 600$.Summing these: $180 + 600 + 600 = 1380$.

The number of integers greater than $3000$ that can be formed by using any number of digits from the set $\{0, 1, 2, 3, 4, 5\}$ without repetition is:

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