How many numbers lying between 999 and 10000 | vedclass

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(C) The numbers lying between $999$ and $10000$ are all $4$-digit numbers.We have $6$ available digits: ${0, 2, 3, 6, 7, 8}$.Total $4$-digit numbers t

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$300$

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(C) The numbers lying between $999$ and $10000$ are all $4$-digit numbers.We have $6$ available digits: ${0, 2, 3, 6, 7, 8}$.Total $4$-digit numbers that can be formed using these $6$ digits (including those starting with $0$) is given by $^6P_4 = 6 	imes 5 	imes 4 	imes 3 = 360$.However,a $4$-digit number cannot start with $0$. If the first digit is $0$,the remaining $3$ places can be filled by the remaining $5$ digits in $^5P_3$ ways.Number of such cases $= ^5P_3 = 5 	imes 4 	imes 3 = 60$.Therefore,the required number of $4$-digit numbers $= 360 - 60 = 300$.

How many numbers lying between $999$ and $10000$ can be formed with the help of the digits $0, 2, 3, 6, 7, 8$ when the digits are not to be repeated?

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