Probablility Question

An increasing list of two digit positive integers is formed that:

- each integer in the list uses digits from {1,2,3,4,5,6}
- each of the integers on the list has the property that its units digit is greater than its ten digit, and
- each of the digits {1,2,3,4,5,6} appears in exactly three of the integers in the list.

How many different lists are possible?

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Since the are 6 digits and each can be used three times, the list should have nine two digit numbers.

So there are 15 posible numbers for the list: 12, 23, 34, 45, 56
13, 24, 35, 46
14, 25, 36
15, 26
16

So from the 15 numbers I need to select 9, or I need to remove 6.
The first number has to be either a 12, 13 or 14 because there needs to be three "1's"

Pick 2 Pick 2 Pick 4
12 - [13,14,15,16] [23,24,25,26] [34,35,36,45,46,56]

Pick 2 Pick 3 Pick 3
13 - [13,14,15,16] [23,24,25,26] [34,35,36,45,46,56]
14 - [13,14,15,16] [23,24,25,26] [34,35,36,45,46,56]

Using the information above I am able to create the diagram above.
The problem is I cant use combinations for the problem because each digit can be used only 3 times.

Another way I could think of the problem is:

Remove 2 Remove 2 Remove 2
12 - [13,14,15,16] [23,24,25,26] [34,35,36,45,46,56]

Remove 2 Remove 1 Remove 3
13 - [13,14,15,16] [23,24,25,26] [34,35,36,45,46,56]
14 - [13,14,15,16] [23,24,25,26] [34,35,36,45,46,56]

This way is easier to do because I need to take less numbers, but combinations still wont work because again each digit can be used only 3 times.

I can't think of any other way to do this problem. Could someone help me?