SOLUTION: with the digits 1,2,3,4 and 5, how many five - digit positive inegers can be formed if the only the 3 can be repeated any number of times?

If there is exactly 1 3,

We pick the 3's position any of C(5,1) ways, and
We position the other 4 digits in P(4,4) ways. 
That's C(5,1)·P(4,4) ways.

If there are exactly 2 3's,

We pick the 3's position any of C(5,2) ways, and
We position the other 3 digits left to right in P(4,3) ways.
That's C(5,2)·P(4,3) ways.


If there are exactly 3 3's,

We pick the 3's position any of C(5,3) ways, and
We position the other 2 digits left to right in P(4,2) ways.
That's C(5,3)·P(4,2) ways.

If there are exactly 4 3's,

We pick the 3's position any of C(5,4) ways, and
We position the other 1 digit in P(4,1) ways.
That's C(5,4)·P(4,1) ways.

If there are exactly 5 3's, 
We pick the 3's position any of C(5,5) ways, and
We position the other 0 digits in P(4,0) ways.
That's C(5,5)·P(4,0) ways.

[Incidentally if there are exactly 5 3's, that works out to
be only one way, 33333.]

Answer: 

C(5,1)·P(4,4) + C(5,2)·P(4,3) + C(5,3)·P(4,2) + C(5,4)·P(4,1) + C(5,5)·P(4,0) =

5·24 + 10·24 + 10·12 + 5·4 + 1·1 = 120 + 240 + 120 + 20 + 1 = 501 ways

Edwin