The total number of three-digit numbers,where | vedclass

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(C) To form a three-digit number with exactly one digit repeated twice,we consider the following cases:Case $1$: The repeated digit is $0$.The number

Vedclass · Accepted Answer

$243$

Vedclass · Answer

(C) To form a three-digit number with exactly one digit repeated twice,we consider the following cases:Case $1$: The repeated digit is $0$.The number is of the form $00x$ or $0x0$ or $x00$. Since it is a three-digit number,the first digit cannot be $0$. Thus,the only possible forms are $x00$ where $x \in \{1, 2, \dots, 9\}$. There are $9$ such numbers.Case $2$: The repeated digit is non-zero (let it be $d \in \{1, 2, \dots, 9\}$).There are $9$ choices for the repeated digit $d$. The third digit $x$ can be any of the remaining $9$ digits (including $0$).If the number is $ddx$,$x$ can be any of the $9$ digits (excluding $d$),giving $9 	imes 8 = 72$ numbers.If the number is $dxd$,$x$ can be any of the $9$ digits (excluding $d$),giving $9 	imes 8 = 72$ numbers.If the number is $xdd$,$x$ can be any of the $8$ digits (excluding $d$ and $0$),giving $9 	imes 8 = 72$ numbers.Total for non-zero repeated digits $= 72 + 72 + 72 = 216 + 18 = 234$ (adjusting for $0$ inclusion).Alternatively: Total numbers with exactly two digits same $= 9 	imes 9 	imes 3 = 243$ (considering positions and digit selection).Total $= 243$.

The total number of three-digit numbers,where exactly one digit is repeated two times,is

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