The number of integers greater than 6000 that | vedclass

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(C) To form an integer greater than $6000$ using the digits ${3, 5, 6, 7, 8}$ without repetition,we consider two cases:Case $1$: $5$-digit numbers.Sin

Vedclass · Accepted Answer

$192$

Vedclass · Answer

(C) To form an integer greater than $6000$ using the digits ${3, 5, 6, 7, 8}$ without repetition,we consider two cases:Case $1$: $5$-digit numbers.Since all $5$ digits are available and we need to form a $5$-digit number,the total number of arrangements is $5! = 5 	imes 4 	imes 3 	imes 2 	imes 1 = 120$.Case $2$: $4$-digit numbers.For a $4$-digit number to be greater than $6000$,the first digit (thousands place) must be $6, 7,$ or $8$. Thus,there are $3$ choices for the first digit.The remaining $3$ positions can be filled by the remaining $4$ digits in $P(4, 3) = 4 	imes 3 	imes 2 = 24$ ways.Total $4$-digit numbers $= 3 	imes 24 = 72$.Total integers $= 120 + 72 = 192$.

The number of integers greater than $6000$ that can be formed using the digits $3, 5, 6, 7,$ and $8$ without repetition is:

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