How many eight-digit serial numbers contain three of one digit, two of another digit, two of yet another digit, and one of still another digit?

It is easier to understand the problem if you create a number based on the criteria and then perform the computations. I am going to choose: 111 22 33 4

There are 10 options for the first "1" and only 1 option for the other two 1's

There are 9 remaining options for the first "2" and only 1 option for the other 2

There are 8 remaining options for the first "3" and only 1 option for the other 3

There are 7 remaining options for the "4"

10 x 1 x 1 x 9 x 1 x 8 x 1 x 7

10 x 9 x 8 x 7 = 5,040

Answer: 5,040