The number of integers greater than 6000 that | vedclass

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(B) To form an integer greater than $6000$,we can form $4$-digit or $5$-digit numbers.Case $1$: $4$-digit numbers greater than $6000$.The first digit

Vedclass · Accepted Answer

$840$

Vedclass · Answer

(B) To form an integer greater than $6000$,we can form $4$-digit or $5$-digit numbers.Case $1$: $4$-digit numbers greater than $6000$.The first digit can be $6, 7, 8,$ or $9$ ($4$ 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 $4$-digit numbers $= 4 	imes 60 = 240$.Case $2$: $5$-digit numbers.Since all $5$-digit numbers formed by these digits are greater than $6000$,we arrange $5$ distinct digits in $5$ positions.Total $5$-digit numbers $= 5! = 120$.Total integers $= 240 + 120 = 360$.Wait,re-evaluating the question: The digits are $0, 5, 6, 7, 8, 9$ (total $6$ digits).For $4$-digit numbers starting with $6, 7, 8, 9$:First digit: $4$ choices.Second digit: $5$ choices.Third digit: $4$ choices.Fourth digit: $3$ choices.Total $= 4 	imes 5 	imes 4 	imes 3 = 240$.For $5$-digit numbers:First digit cannot be $0$ ($5$ choices).Remaining $4$ positions: $P(5, 4) = 5 	imes 4 	imes 3 	imes 2 = 120$.Total $= 5 	imes 120 = 600$.Total $= 240 + 600 = 840$.

The number of integers greater than $6000$ that can be formed by using the digits $0, 5, 6, 7, 8,$ and $9$ without repetition is

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