jmc

Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113,B=13,C=31$, then the string is $$1131331.$$ Since the strings have seven digits and three threes, there are $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ arrangements of all such strings. In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups. Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to $$x+y+z=7,x,y,z>0.$$ This gives us $$\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$$ ways by balls and urns. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$. Thus, each arrangement has$$\left(\genfrac{}{}{0ex}{}{6}{2}\right)-3=12$$ways per arrangement, and there are $12\times 35=\boxed{420}$ ways.