Permutation-Combination
Aptitude

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Q.

How many numbers are there between 100 and 1000 such that at least one of their digits is 6?

 A.

200

 B.

225

 C.

252

 D.

120

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Solution:
Option(C) is correct

numbers between 100 and 1000:

$= 900$ 

numbers between 100 and 1000 which do not have digit 6 in any place:

$= 8 × 9 × 9$

$= 648$

Unit digit could take any value of the 9 values (0 to 9, except 6) 

Tens Digit could take any value of the 9 values (0 to 9, except 6) 

Hundreds digit could take any value of the 8 values (1 to 9, except 6) 

numbers between 100 and 1000 which have at least one digit as 6:

$= 900 - 648$

$= \textbf{252}$


(1) Comment(s)


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The answer is actually 253. You included 100 when subtracting (900 total between) but you didn't subtract 1 from numbers between 100 and 1000 for when your math lands on 1 x 0 x 0.