How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?	Select the correct option
A) 400  B) 1728  C) 108  D) 216  E) 432
The number of ways u can chose two digits from 9 digits is 9C2=(9*8)/2.   And the no of ways u can arrange two digits like asked is 4C2=(4*3)/2.   total ways is   36*6= 
<gadget id="calculator">  36*6</gadget>
<output>216</output>
216.   Answer : 216
<result>D</result>
How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?