The number of natural numbers less than 500 i | vedclass

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Question

(C) To find the number of natural numbers less than $500$ with no repeated digits,we consider $1, 2,$ and $3$-digit numbers separately.$1$. $1$-digit

Vedclass · Accepted Answer

$378$

Vedclass · Answer

(C) To find the number of natural numbers less than $500$ with no repeated digits,we consider $1, 2,$ and $3$-digit numbers separately.$1$. $1$-digit numbers: The possible numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9$. Total = $9$.$2$. $2$-digit numbers: The first digit can be any of $9$ digits $(1-9)$ and the second digit can be any of the remaining $9$ digits (including $0$). Total = $9 	imes 9 = 81$.$3$. $3$-digit numbers less than $500$: The hundreds place can be filled by $1, 2, 3,$ or $4$ ($4$ ways). The tens place can be filled by any of the remaining $9$ digits,and the units place by any of the remaining $8$ digits. Total = $4 	imes 9 	imes 8 = 288$.Total count = $9 + 81 + 288 = 378$.

The number of natural numbers less than $500$ in which no two digits are repeated is

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