Of the three-digit integers greater than 900, how many have two digits that are equal to each other and the remaining digit different from the other two? Choose one: A) 94 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=95   94nswer is 94 ('cause 700 can't be included)   94nswer (94).
<result>A</result>
Of the three-digit integers greater than 900, how many have two digits that are equal to each other and the remaining digit different from the other two?