Let $n>0$ be an integer. Let also $a_j$ be some integer in the set $\{0,1,\ldots,\binom{n}{j}\}$ for all $j=0,1,\ldots,n$.
Then, how many integers can be written in the form $$2^n a_n+2^{n-1}\cdot 3 a_{n-1}+\ldots+2\cdot3^{n-1}a_1+3^n a_0?$$
A Note: Suppose that two integers can be represented in the same way in the above form. Then there exist integers $a_j$ in the same sets such that, for suitable choices of the signs, we have $$\pm 2^n a_n\pm 2^{n-1}\cdot 3 a_{n-1}\pm \ldots\pm 2\cdot3^{n-1}a_1\pm 3^n a_0=0.$$