The number of 4-digit numbers that can be for | vedclass

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(D) To find the number of $4$-digit numbers containing at least one $1$,we use the complement method:Total $4$-digit numbers that can be formed using

Vedclass · Accepted Answer

$1876$

Vedclass · Answer

(D) To find the number of $4$-digit numbers containing at least one $1$,we use the complement method:Total $4$-digit numbers that can be formed using the digits ${0, 1, 2, 3, 4, 5, 6, 7}$ (without repetition):The first digit cannot be $0$,so there are $7$ choices for the first place $(1-7)$.The remaining $3$ places can be filled by the remaining $7$ digits in $^7P_3$ ways.Total $4$-digit numbers $= 7 	imes 7 	imes 6 	imes 5 = 1470$.Now,find the number of $4$-digit numbers that do $NOT$ contain the digit $1$:The digits available are ${0, 2, 3, 4, 5, 6, 7}$ (total $7$ digits).The first digit cannot be $0$ or $1$,so there are $6$ choices $(2, 3, 4, 5, 6, 7)$.The remaining $3$ places can be filled by the remaining $6$ digits in $^6P_3$ ways.Numbers without $1 = 6 	imes 6 	imes 5 	imes 4 = 720$.Therefore,the number of $4$-digit numbers containing at least one $1 = 1470 - 720 = 750$.

The number of $4$-digit numbers that can be formed from the digits $0, 1, 2, 3, 4, 5, 6, 7$ such that each number contains the digit $1$ is:

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