How many 4-digit numbers can be formed by using the digits 0

How many 4-digit numbers can be formed by using the digits 0-9, so that the numbers contains exactly 3 distinct digits?

(A) 1944
(B) 3240
(C) 3850
(D) 3888
(E) 4216


So I did this a slightly different way - which I think is easier and can be done in less than a minute:


Total number of 4 digit numbers that contain exactly 3 distinct digits (including digits starting with zero):

10c3 = number of ways to select 3 distinct digits from the 10 available (0-9)

(4!/2!) * 3 = number of ways to arrange these 3 digits where one is repeated (similar to a word problem). For example ABCC, ACBC, CCAB.... there are 4!/2! ways to arrange, and 3 ways in which you could duplicate the number (AA,BB,CC).

Total = 10c3 x (4!/2!)*3 = 120 * 12 * 3 = 4,320

Note: This includes invalid 4 digits, e.g 0031, 0131, 0910 etc. where it starts with zero or double zero.


Invalid 4 digits starting with zero:

The number of these is 4320/10 (as there is a 1/10 chance that the number starts with a 0).


Total - Invalid "zero" 4 digit numbers = 4,320 - 432 = 3,888

Choose D.