A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?  Pick: A) 72 B) 144 C) 216 D) 283 E) 300
here is the answer in pure probability terms.   let the 3 digit number be XYZ   now none of them contain zero --&gt; they can vary from 1 - 9   so lets take the value of X, X can be from 1 - 9, hence X can be selected in 9 ways   since Y =X, after selecting X you can select Y in 1 way only   Z can be an selected from 1 -9 (minus 1 because it cannot be the same as X) in exactly 8 ways   hence total number of ways of selecting the number is   9 * 8= 
<gadget id="calculator">  9 * 8</gadget>
<output>72</output>
72.   BUT we are asked for all possible numbers that can be formed.   So XYZ ( or XXZ in this case) can be moved around in 3 ways   XXZ   XZX   ZXX   hence the total number of numbers is 72 *3= 
<gadget id="calculator">72 *3</gadget>
<output>216</output>
216   ANSWER:216
<result>C</result>
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?