Let $N^{(k)}_{a,b,c}$ be the number of possible unnumbered layouts of the last $k$ columns, given that there are $a$ numbers left to assign in the first row, $b$ in the second row, and $c$ in the third row. Given the rules (at least one number per column), you have the recursion
$$
\begin{eqnarray}
N^{(k)}_{a,b,c}&=&N^{(k-1)}_{a-1,b-1,c-1}+N^{(k-1)}_{a-1,b-1,c}+N^{(k-1)}_{a-1,b,c-1}+N^{(k-1)}_{a,b-1,c-1} + N^{(k-1)}_{a-1,b,c}+N^{(k-1)}_{a,b-1,c}+N^{(k-1)}_{a,b,c-1},
\end{eqnarray}
$$
with the boundary conditions that $N^{(0)}_{0,0,0}=1$ and $N^{(k)}_{a,b,c}=0$ unless all indices are between $0$ and $k$. You want to find $N^{(9)}_{5,5,5}$. The following Python function performs the calculation:
def N(a,b,c,k,cache={(0,0,0,0):1}):
if min(a,b,c)<0 or max(a,b,c)>k:
return 0
if not cache.has_key((a,b,c,k)):
val = N(a-1,b-1,c-1,k-1)
val += N(a-1,b-1,c,k-1) + N(a-1,b,c-1,k-1) + N(a,b-1,c-1,k-1)
val += N(a-1,b,c,k-1) + N(a,b-1,c,k-1) + N(a,b,c-1,k-1)
cache[(a,b,c,k)] = val
return cache[(a,b,c,k)]
N(5,5,5,9)
> 735210
The result is below the upper bound of ${{9}\choose{5}}^3=126^3=2000376$ obtained by allowing all-blank columns, as it should be.
For numbered layouts, the recursion is slightly different, because the number of ways to choose the numbers depends on the column. Here, let $m_k$ be the number of allowed numbers in the $k$-th column from the end: $m_1=11$, $m_9=9$, and $m_k=10$ otherwise. In a column with no blanks, there are ${m_k}\choose{3}$ ways to choose the numbers; with one blank, ${m_k}\choose{2}$; and with two blanks, $m_k$. The recursion is therefore
$$
\begin{eqnarray}
M^{(k)}_{a,b,c}&=&{{m_k}\choose{3}}M^{(k-1)}_{a-1,b-1,c-1}+ {{m_k}\choose{2}}\left(M^{(k-1)}_{a-1,b-1,c}+M^{(k-1)}_{a-1,b,c-1}+M^{(k-1)}_{a,b-1,c-1}\right) \\ && + m_k\left(M^{(k-1)}_{a-1,b,c}+M^{(k-1)}_{a,b-1,c}+M^{(k-1)}_{a,b,c-1}\right).
\end{eqnarray}
$$
Not unexpectedly, the result here is much larger:
$$
M^{(9)}_{5,5,5} = 3669688706217187500.
$$
As a sanity check on this result, one might consider the average branching factor for the $15$ numbers in an average unnumbered layout:
$$
\left(\frac{M^{(9)}_{5,5,5}}{N^{(9)}_{5,5,5}}\right)^{\frac{1}{15}} \approx 7.023.
$$
Since in a typical row (with $m_k=10$) the possible outcomes have average branching factors ${{10}\choose{3}}^{1/3}\approx 4.93$, ${{10}\choose{2}}^{1/2}\approx 6.71$, and $10$, this seems very reasonable.
Note that we can also enumerate the possible unnumbered layouts in an entirely different way, using inclusion-exclusion. The basic idea is to count all the ways of placing $5$ numbers in each of the three rows, then remove the cases with an all-blank column, then add back in the cases with two all-blank columns, and so on:
$$
\begin{eqnarray}
N^{(9)}_{5,5,5} &=& {{9}\choose{5}}^3 - {{9}\choose{1}}{{8}\choose{5}}^3 + {{9}\choose{2}}{{7}\choose{5}}^3 - {{9}\choose{3}}{{6}\choose{5}}^3 + {{9}\choose{4}} \\
&=& 735210,
\end{eqnarray}
$$
as above. This is an elegant solution to the unnumbered case, but I do not see a way to extend it to the numbered layouts.
Best Answer
There are four complete columns of 5 and one with only 4 numbers. For column 1 there are: $$5! {15 \choose 5}$$ Since there are 5 numbers out of a possible 15 used in the column and these can be permuted in any order (in 5! ways). This also holds for the other 3 complete columns. The same method works for the column of 4 giving instead: $$4! {15 \choose 4} $$
Therefore the total answer is: $$ \left (5! {15 \choose 5} \right)^4 4! {15 \choose 4} = 552446474061128648601600000 $$