I try here to expose a way to build solutions manually. 

For easiness I take B = 3. Then, obviously C begins (left) by 1 or 2, so we have to reserve that digit for that position (C = 1abcdefgh or 2abcdefgh). 

When A is multiplied by 3, each digit of A receive from the contiguous-right digit an addition (0, 1 or 2 units) and transfer to the contiguous-left another addition (=0, 1 or 2 units). In particular: 1,2 when multiplied by 3 do not transfer additions; 4,5,6 transfer one unit; 7,8,9 transfer two units. 

So, when a digit of A is multiplied by 3, the corresponding digit in the product C would result in three possibles digits (only two for number 6, as we don't allow 0 in C). 

For example: digit 7 of A ends in 1 when multiplied by 3. But its contiguous number on the right, if existent, would transfer to it an amount of 0,1 or 2 units, so the corresponding digit in C will be 1,2 or 3. This flexibility of these additions is what give way to repetitions in C.

To assure that C has each digit just one time we can:

1. compile a table with the three possibles values for each A-digit. 

   A= 7  4   1   8   5   2  9  6

C      

1      x   

2      x   x   

3      x   x   x  

4           x   x  x  

5                x  x  x  

6                    x  x  x  

7                        x  x  x

8                            x  x  x

9                                x  x

2. choose a disposition, taking care that each row and column (except row 1, as C begins by 1) has just one element chosen (see below). 

3. Now that the nine digits in C (the product) have been chosen, we have to determine the order of the digits in C. 

To do this I have to follow a coherent path that begins with a circle and goes circle by circle to the ninth one. 

The only possible circles to begin with are under the A-digits 1, 5 or 6: this is necessary as the first-right A-digit multiplied (-------*) do not receive transfers from other digits, but only can add to the next one). (Also I should should not finish with circle on row two, as it would lead to a 8-digits in C).

Let us try: 

I choose to begin with the circle under A-digit 1 (underlined). It is not going to transfer (as 1*3=3), so I go to other circle with has no transfer (they are under A-digits 5 or 6), I choose 5. It transfers 1 (as 5*3=15), so I have to go to a circle having addition of 1.  They are under A-digits 2 or 7. I choose 2. It does not transfer (as 3*2=06 and it has receive only one, total will be: 07). So have to go to a circle receiving no additions: only remain circle under A-digit 6. 6 transfers 1, so I go to 7, whose circle has the addition of one. 7 transfers 2 (7+3=21), so it send to 8 or to 9. Choose 8, transfers 2; send me to 9; transfers 2; and send me to 4 which is the last circle. 

A: 1 => 5 => 2 => 6 => 7 => 8 => 9 => 4

  49876251*3=149628753