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The number of integer’s n with $100 \leqslant n \leqslant 999$and containing at most two distinct digits is:
$
  {\text{a}}{\text{. }}252 \\
  {\text{b}}{\text{. }}280 \\
  {\text{c}}{\text{. }}324 \\
  {\text{d}}{\text{. }}360 \\
$

Answer
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Hint: - Calculate total numbers having distinct digits then subtract this from total numbers.

Total three digit numbers between $100 \leqslant n \leqslant 999$
$999 - 99 = 900$, because 999 and 100 are included.
Now, a three digit number is to be formed from the digits $0,1,2,3,4,5,6,7,8,9$
$ \bullet \bullet \bullet $
Since the left most place i.e. hundred’s place cannot have zero.
So, there are 9 ways to fill hundred’s place.
Since, we consider the number with distinct digits, therefore repletion is not allowed, so, ten’s place can be filled by 9 remaining ways.
So, ten’s place can be filled in 9 ways.
Similarly, to fill the unit's place, we have 8 digits remaining.
So, the unit's place can be filled by 8 ways.
So, the required number of ways in which three distinct digit number can be formed are
$9 \times 9 \times 8 = 648$
So, the numbers having all the distinct digits $ = 648$
Thus, the remaining numbers containing at most two distinct digits $ = $ total numbers $ - $numbers having all the distinct digits.
$ = 900 - 648 = 252$
Hence, option $a$ is correct.

Note: - In such types of questions first calculate the total numbers, then calculate the total numbers having distinct digits using the procedure which is stated above, then subtract these values we will get the required answer.