The number of integers lying between 1000 and | vedclass

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(C) We need to form a $4$-digit number using the digits $3$ and $7$ exactly once. The remaining $2$ positions can be filled by any of the remaining $8

Vedclass · Accepted Answer

$720$

Vedclass · Answer

(C) We need to form a $4$-digit number using the digits $3$ and $7$ exactly once. The remaining $2$ positions can be filled by any of the remaining $8$ digits $(0, 1, 2, 4, 5, 6, 8, 9)$ without repetition.Total ways to choose $2$ positions for $3$ and $7$ out of $4$ is $^4C_2 = 6$. The digits $3$ and $7$ can be arranged in $2! = 2$ ways.The remaining $2$ positions can be filled by $8$ digits in $P(8, 2) = 8 	imes 7 = 56$ ways.Total numbers including those starting with $0$ is $6 	imes 2 	imes 56 = 672$.Now,subtract the cases where the number starts with $0$ (i.e.,$3$-digit numbers).If the first digit is $0$,we choose $2$ positions for $3$ and $7$ from the remaining $3$ positions in $^3C_2 = 3$ ways. They can be arranged in $2! = 2$ ways. The last remaining position can be filled by any of the $7$ remaining digits in $7$ ways.Numbers starting with $0 = 3 	imes 2 	imes 7 = 42$.Required number $= 672 - 42 = 630$.

The number of integers lying between $1000$ and $10000$ such that every number contains the digits $3$ and $7$ exactly once without repetition is:

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