5-digit numbers are to be formed using 2, 3, | vedclass

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(D) The total number of $5$-digit numbers that can be formed using the digits ${2, 3, 5, 7, 9}$ without repetition is $5! = 120$.Since all these digit

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$5:3$

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(D) The total number of $5$-digit numbers that can be formed using the digits ${2, 3, 5, 7, 9}$ without repetition is $5! = 120$.Since all these digits are greater than $2$,any $5$-digit number formed using these digits will always be greater than $20000$. Thus,$p = 120$.For $q$,the numbers must lie between $30000$ and $90000$. This means the first digit (ten-thousands place) can only be $3, 5,$ or $7$.There are $3$ choices for the first digit.The remaining $4$ positions can be filled by the remaining $4$ digits in $4!$ ways.Therefore,$q = 3 	imes 4! = 3 	imes 24 = 72$.The ratio $p : q = 120 : 72$.Dividing both by $24$,we get $p : q = 5 : 3$.

$5$-digit numbers are to be formed using $2, 3, 5, 7, 9$ without repeating the digits. If $p$ is the number of such numbers that exceed $20000$ and $q$ is the number of those that lie between $30000$ and $90000$,then $p : q$ is

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