Of the two-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?  Select the correct option.
A) 24
B) 9
C) 72
D) 144
E) 216
Of the two-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?   A. 24   B. 36   C. 72   D. 144   E. 216   Choosing the digit for X - 9 ways;   # of permutations of 3 digits in XX -2!/2!   Total: 9*2!/2!=9.   Answer: 9.  9*2 /2= 
<gadget id="calculator">  9*2 /2</gadget>
<output>9</output>
9.   Answer: 9.
<result>B</result>
Of the two-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?