Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two? Choose the correct option:
A) 24 B) 36 C) 72 D) 1 E) 216
Assume: a,b,c is the digit and a,b,c not= 0   Thus a,b,c could be 1, 2, 3, 4, 5, 6, 7, 8, and 9   Count the possible way to get three digit number abc.   a = b, and c must be different from a, b   Thus, there are 3 possible ways of digit arrangement: aac, aca, caa   Case I: aac   =&gt; (digit 1st) x (digit 2nd) x (digit 3rd)   =&gt; 9 x 1 x 8 {pick any number from group = 9 possible ways} x {pick number the same as the first pick = 1 way} x {pick any number from the rest = 8 possible ways}   = 9 x 1 x 8 = 72 possible ways   Case II: aca   =&gt; same as case I you have 72 possible ways   Case III: caa   =&gt; same as case I you have 72 possible ways   total of this set of number = 72 + 72 + 72= 
<gadget id="calculator">72 + 72 + 72</gadget>
<output>216</output>
216   216) is the answer
<result>E</result>
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?