SOLUTION: How many 8 digit numbers are there with 3 different digits, 1 appearing twice and the other two appearing three times each? Assume zero cannot be used at all.

There are 9C3 ways to choose the 3 digits.

Of those 3 choose the digit to be used only twice 3C1 ways. 

There are 8 places in an 8-digit number for the digits to go.

Choose the 2 places for the digit to be used twice in 8C2 ways.
That leaves 6 places for the larger of the 2 remaining digits.
Choose the 3 places for the larger remaining digit to go in 6C3 ways.
That leaves 3 places for the smaller remaining digit.
Choose the places for the smallest digits to go in 3C3 or 1 way. 
(Only 1 way, because there are only 3 places left for the 
smallest digit).

Answer: (9C3)(3C1)(8C2)(6C3)(3C3) = 141120 ways 

Edwin