There are two ways of doing this. One is Ross Millikan's: you will make ten "up" moves, and 20 "right" moves; the only question is which order you make them in. Imagine placing the "right" moves on a row; now you need to decide where to do the "up" moves: you do so by inserting them "in between" (or before, or after) the "right" moves. So you need to choose ten places to put "up" moves: there are 21 locations for them (nineteeen in between the "right" moves, one before all of them, one after), and you are allowed to choose the same location more than once.
This is a combinations-with-repetitions: the formula is $\binom{n+r-1}{r}$, where you have $n$ possibilities, and must make $r$ choices with repetitions allowed. In this case, $n=21$, $r=10$, so you get $\binom{30}{10}$.
There is another way of doing it, which is more graphical. I'll do it with a 4 by 3 array so you see how it works. You have this array:
$$\begin{array}{cccc}
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot
\end{array}$$
Now, you start at the bottom left, so there is only one way to get there; we put a $1$ next to it.
$$\begin{array}{llll}
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot\;1& \cdot & \cdot & \cdot
\end{array}$$
Then, you can go either up or right; there is only one way to get to those points (via the first move); we put a $1$ next to them:
$$\begin{array}{llll}
\cdot & \cdot & \cdot & \cdot\\
\cdot\;1 & \cdot & \cdot & \cdot\\
\cdot\;1& \cdot\;1 & \cdot & \cdot
\end{array}$$
Now: to get to $(1,1)$, you can either get to it from $(1,0)$ or from $(0,1)$; since there is only one way to get to each of those, there are two ways to get to $(1,1)$. On the other hand, only one way to get to $(2,0)$ or to $(0,2)$:
$$\begin{array}{llll}
\cdot\;1 & \cdot & \cdot & \cdot\\
\cdot\;1 & \cdot\;2 & \cdot & \cdot\\
\cdot\;1& \cdot\;1 & \cdot\;1 & \cdot
\end{array}$$
Next: to get to $(1,2)$, you can arrive either from $(0,2)$ (one way of being there), or from $(1,1)$ (two ways of getting there); so in total, three ways. Likewise, you have three ways to get to $(2,1)$, because you can either go up from $(2,0)$, and there is only one way to do all of that, or you can go right from $(1,1)$ (and there are two ways of doing that, corresponding to the two ways there are to get to $(1,1)$; so we have:
$$\begin{array}{llll}
\cdot \;1 &\cdot\;3 & \cdot & \cdot\\
\cdot\;1 & \cdot\;2 & \cdot\;3 & \cdot\\
\cdot\;1& \cdot\;1 & \cdot\;1 & \cdot
\end{array}$$
Continuing this way, we get:
$$\begin{array}{llll}
\cdot\;1 & \cdot\;3 & \cdot\;6 & \cdot\;10\\
\cdot\;1 & \cdot\;2 & \cdot\;3 & \cdot\;4\\
\cdot\;1 & \cdot\;1 & \cdot\;1 & \cdot\;1
\end{array}$$
So there are $10$ ways to get to the top right corner in the 4 by 3 case.
You may even recognize that these numbers are just Pascal's triangle lying on its side! Well, there is a combinatorial formula for the entries of Pascal's triangle: the $r$th entry in the $m$th row corresponds to the coefficient of $a^{m-r}b^{r-1}$ in the binomial expansion of $(a+b)^{m-1}$, so it equals $\binom{m-1}{r-1}$. To figure out the entry that corresponds to the top right corner, note that you go "down" one row for each position on the $x$-axis, and another one for each step up. So here we have gone to the 4th row on the horizontal steps, and then to the 6th row on the by going up twice. Each step up is a move "right" on the row. So with a 4 by 3, we are in row $4+(3-1)=6$, and we are in position $3$ of that row. According to the formula above, that corresponds to $\binom{4+2-1}{3-1}=\binom{5}{2}$. This corresponds to the need to make $3$ moves right and two moves up, so we need choose where to place the two up moves among the three right moves; there are four places to put them in (before the three, after the three, or in the two spaces in between), so the formula I gave above gives this answer as well.
Let $d$ be the number of diagonal moves; clearly $0\le d\le 10$. The $d$ diagonal moves account altogether for $d$ steps to the right and $d$ steps up, so he must also make $10-d$ moves to the right and $10-d$ moves up. That’s a total of $20-2d$ orthogonal moves, half to the right and half up, which may be taken in any order, for a grand total of $20-2d+d=20-d$ moves.
There are $\binom{20-d}d$ ways to choose which moves are diagonal, and then there are $\binom{20-2d}{10-d}$ ways to choose which of the remaining moves are up, so there are
$\binom{20-d}d\binom{20-2d}{10-d}$ paths with $d$ diagonal moves and a grand total of
$$\sum_{d=0}^{10}\binom{20-d}d\binom{20-2d}{10-d}=\sum_{k=0}^{10}\binom{10+k}{10-k}\binom{2k}k\;.$$
You could calculate this longhand, or you could calculate the corresponding figure for smaller numbers than $10$: for $0$ I get $1$, for $1$ I get $\binom11\binom00+\binom20\binom21=3$, for $2$ I get $$\binom22\binom00+\binom31\binom21+\binom40\binom42=13\;,$$ and for $3$ I get $$\binom33\binom00+\binom42\binom21+\binom51\binom42+\binom60\binom63=1+12+30+20=63\;,$$ which is enough to justify trying the On-Line Encyclopedia of Integer Sequences. It reports $14$ matches, of which the first, A001850, gives the central Delannoy numbers, described as the number of paths from $(0,0)$ to $(n,n)$ in an $n\times n$ grid using only steps North, Northeast and East, which is exactly what you want. The tenth number is listed as $8,097,453$.
Added: The sequence UUDDUD, which I’ll call $S$, has no net effect and may be inserted at any point before the last two moves that increase the $y$-coordinate. As before, suppose that there are $d$ diagonal moves, so that there are $10-d$ up and $10-d$ right moves, not including $S$. Still not counting $S$, there are $10$ moves that increase the $y$-coordinate; call them $m_1,\dots,m_{10}$ in order. They may come in any order, and there are $\binom{10}d$ ways to choose which of them are the diagonal moves. $S$ may come before $m_1$ or between $m_k$ and $m_{k+1}$ for $k=1,\dots,8$, so there are $9$ places to insert $S$ into any of those $\binom{10}d$ possible sequences, for a total of $9\binom{10}d$ sequences of $S$ and the $10$ moves that increase the $y$-coordinate.
The remaining $10-d$ moves may be placed anywhere in the string. This is a standard stars and bars problem: there are $12$ possible places to put the moves, and $10-d$ indistinguishable moves to be placed, for a total of $\binom{10-d+11}{11}=\binom{21-d}{11}$ arrangements. The grand total is therefore
$$9\sum_{d=0}^{10}\binom{10}d\binom{21-d}{11}$$
paths. If I did the arithmetic correctly, this comes to $119,574,387$ paths.
Best Answer
I don't know of a combinatorial solution, but I do know a trick...
The number of $n$-step walks starting at $(x,y)$ that don't leave the grid is obviously the total number of $n$-step walks ($4^n$) times the probability $p_n(x,y)$ that a random $n$-step walk started at $(x,y)$ doesn't leave the grid.
What is that probability? Obviously, $p_0(x,y) = \chi(x,y)$, where $\chi(x,y)$ equals $1$ inside the grid and $0$ on the points surrounding it. For higher $n$, we have the recurrence relation $$p_{n+1}(x,y) = \chi(x,y) \frac{p_n(x-1,y) + p_n(x+1,y) + p_n(x,y-1) + p_n(x,y+1)}{4}.$$
Except for the $\chi(x,y)$ term, this is almost a convolution, and by making use of the fact that the grid is rectangular, we can get rid of the unwanted term by extending $p_0$ suitably beyond the edges of the grid.
Here it turns out to be much more convenient to let the grid extend from $(1,1)$ to $(a-1, b-1)$, so I'm going to adopt that convention, and then define $p_0(x,y) = f(x)g(y)$, where $$f(x) = \begin{cases} \phantom{+}0 & \text{if }x = ka, k \in \mathbb Z \\ \phantom{+}1 & \text{if }(2k+0)a < x < (2k+1)a, k \in \mathbb Z \\ -1 & \text{if }(2k+1)a < x < (2k+2)a, k \in \mathbb Z \end{cases}$$ $$g(y) = \begin{cases} \phantom{+}0 & \text{if }y = kb, k \in \mathbb Z \\ \phantom{+}1 & \text{if }(2k+0)b < y < (2k+1)b, k \in \mathbb Z \\ -1 & \text{if }(2k+1)b < y < (2k+2)b, k \in \mathbb Z.\end{cases}$$
It is then not hard to show that, even if we drop the $\chi(x,y)$ term from the recurrence, $p_n(x,y) = 0$ whenever $x$ is divisible by $a$ or $y$ is divisible by $b$. Thus, we can calculate $p_{n+1}$ for $n \ge 0$ via the convolution $p_{n+1} = p_n * K$, i.e. $$p_{n+1}(x,y) = \sum_{u,v \in \mathbb Z} p_n(x-u,y-v) K(u,v),$$ where the kernel $K$ is given by $K(1,0) = K(0,1) = K(-1,0) = K(0,-1) = \frac 1 4$ and $K(u,v) = 0$ for all other values of $u$ and $v$.
Why would we want to do that? Well, it just happens that convolutions of discrete periodic functions can be efficiently calculated using discrete Fourier transforms. In particular, letting $\tilde p_n$ and $\tilde K$ denote the Fourier transforms of $p_n$ and $K$, we have $$\tilde p_n = \tilde p_{n-1} \tilde K = \tilde p_0 \tilde K^n,$$ where the multiplication is simply done pointwise.