How many three digit numbers contain the digit 5 at least once?	Options.
A) 52  B) 128  C) 252  D) 648  E) 900
The question is basically asking how many numbers between 100-999 contain at least one digit as 5.   Let ABC be a 3 digit number A can be filled with 9 digits, B can be filled with 10 digits and C can be filled with 10 digits. (because hundredth digit can only be between 1-9= 9 digits)   Therefore total number of ways we can form 3 digit number=9*10*10= 
<gadget id="calculator">9*10*10</gadget>
<output>900</output>
900 ways (100-999)   Number of ways we can form a 3 digit number without the digit 5= 8*9*9= 
<gadget id="calculator">8*9*9</gadget>
<output>648</output>
648 ways   ANSWER:648
<result>D</result>
How many three digit numbers contain the digit 5 at least once?