Counting

A lot of casework, but relatively simple.

We find the digits for each number, find the number of ways to arrange them, to represent the different integers.

Here are the possible cases:

5 = 

C1) 1, 1, 1, 1, 1

C2) 1, 1, 1, 2

C3) 1, 2, 2

C4) 1, 1, 3

C5) 2, 3

For C1, there is 1 way to order this.

For C2, there is \(\frac{4!}{3!1!}\), or 4 ways to order this.

For C3, there is \(\frac{3!}{2!1!} \), or 3 ways to order this.

For C4, there is \(\frac{3!}{2!1!}\), or 3 ways to order this.

For C5, there is 2 ways to order this.

 

There are a total of 1+4+3+3+2 = 13 numbers.