Question

(i) How many numbers are there between $99$ and $1000$ having $7$ in the units place?
(ii) How many numbers are there between $99$ and $1000$ having at least one of their digits is $7$?

• A. $ 90,253 $
• B. $ 90,258 $
• C. $ 90,252 $
• D. $ 90,352 $

Answer 1

Answer: (C)

(i) We know that all these numbers have three digits. $7$ is in the unit's place. The middle digit can be any one of the $10$ digits from $0$ to $9$. The digit in hundred's place can be any one of the $9$ digits from $1$ to $9$. Therefore, by the fundamental principle of counting, there are $10 \times 9 = 90$ numbers between $99 $ and $1000$ having $7$ in the unit's place.
(ii) Total number of $3$ digit numbers having atleast one of their digits as $7$ = (Total numbers of three digit numbers) - (Total number of $3$ digit numbers in which $7$ does not appear at all)
$= (9 \times 10 \times 10) - (8 \times 9 \times 9) = 900 - 648$
$ = 252$.