Solution:

Answer: 3357

From any sequence $a_{1}, a_{2}, \ldots, a_{10}$, construct a sequence $b_{1}, b_{2}, \ldots, b_{9}$, where $b_{i}$ counts the number of times $i$ occurs in the sequence. There is a correspondence from all possible sequences $b_{1}, b_{2}, \ldots, b_{9}$ with at most 3 nonzero terms which add to 10, since any sequence of $a_{1}, a_{2}, \ldots, a_{10}$ will be converted to this form, and from any sequence $b_{1}, b_{2}, \ldots, b_{9}$, we can construct a unique sequence of $a$-s by listing $i$ $b_{i}$ times (for $1 \leq i \leq 9$) in nondecreasing order.

Our goal now is to count the number of possible sequences $b_{1}, b_{2}, \ldots, b_{9}$ meeting our conditions. We casework on the number of nonzero terms in the sequence:

Case 1: The sequence has exactly one nonzero term.

Then exactly one of $b_{1}, b_{2}, \ldots, b_{9}$ is equal to 10, and all the rest are equal to 0. This gives us 9 possible sequences in this case.

Case 2: The sequence has exactly two nonzero terms.

There are $\binom{9}{2} = 36$ ways to choose the two terms $b_{i}, b_{j}$ ($i < j$) which are nonzero. From here, we have 9 choices for the value of $b_{i}$, namely 1 through 9 (since both $b_{i}$ and $b_{j}$ must be nonzero), and $b_{j}$ will be fixed, so this case gives us $36 \cdot 9 = 324$ possible sequences.

Case 3: The sequence has exactly three nonzero terms.

There are $\binom{9}{3} = 84$ ways to choose the three terms $b_{i}, b_{j}, b_{k}$ ($i < j < k$) which are nonzero. Letting $c_{i} = b_{i} - 1$, $c_{j} = b_{j} - 1$, $c_{k} = b_{k} - 1$, we have that $c_{i}, c_{j}, c_{k}$ are nonnegative integers which sum to 7. There are $\binom{9}{2} = 36$ solutions to this equation (consider placing two dividers in the nine spaces between the ten elements), giving $84 \cdot 36 = 3024$ possibilities in this case.

We then have $9 + 324 + 3024 = 3357$ possible sequences.