How many numbers lying between 999 and 10000 | vedclass

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Question

(C) The numbers lying between $999$ and $10000$ are all $4$-digit numbers.The total number of $4$-digit numbers that can be formed using the digits ${

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(C) The numbers lying between $999$ and $10000$ are all $4$-digit numbers.The total number of $4$-digit numbers that can be formed using the digits ${0, 2, 3, 6, 7, 8}$ without repetition is given by $^6P_4 = \frac{6!}{(6-4)!} = 6 	imes 5 	imes 4 	imes 3 = 360$.However,a $4$-digit number cannot begin with the digit $0$. We must exclude those cases where $0$ is in the thousands place.If the thousands place is fixed as $0$,the remaining $3$ places can be filled by the remaining $5$ digits ${2, 3, 6, 7, 8}$ in $^5P_3 = \frac{5!}{(5-3)!} = 5 	imes 4 	imes 3 = 60$ ways.Therefore,the required number of $4$-digit numbers is $360 - 60 = 300$.

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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