Of the three-digit integers greater than 400, how many have two digits that are equal to each other and the remaining digit different from the other two?	Choose the correct choice.
A) 90  B) 82  C) 80  D) 45  E) 36
To satify the given condition,   required no. of cases = total numbers - numbers with all digits different - numbers when all three digits are same,   number greater than 700;   total numbers = 1*10*10= 
<gadget id="calculator">1*10*10</gadget>
<output>100</output>
100   numbers with all digits different= 1*9*8= 
<gadget id="calculator">1*9*8</gadget>
<output>72</output>
72   numbers when all three digits are same (777)= 1   req. = 100- 72 - 1= 
<gadget id="calculator">100- 72 - 1</gadget>
<output>27</output>
27   considering the numbers between 700999= 27*3=37   Answer is 36('cause 700 can't be included)   Answer (36).
<result>E</result>
Of the three-digit integers greater than 400, how many have two digits that are equal to each other and the remaining digit different from the other two?