In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4? Choose one.
A) 5P3/(2!)  B) 4P3  C) 4^3  D) 4P3+3C1*(3!/2!)  E) 60 × 3!
Had it asked for all the different nos instead of 1,1,2,3,4 the ans would be: 5P3/2! = 30   In this solution we have also halved (divided by 2!) the numbers containing 2,3 and 4.   The numbers containing 2,3,4 are 6.   Add  6/2= 
<gadget id="calculator"> 6/2</gadget>
<output>3</output>
3 back to 30 and we get the answer i.e. 30+3=33.   Hence 4P3+3C1*(3!/2!) is the answer
<result>D</result>
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?