different path permutation problem Im having hard time with this question


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*How many different paths in the $xy$-plane are there from $(0,0)$ to $(7,7)$ if a path proceeds either one space to the right $(r)$ or on space upward $(u)$? 
Here's what I did: $14!/7!7!$

*$(2,7)$ to $(9,14)$? what I did: $14!7!7!$

*How many such paths are there from $(a,b)$ to $(c,d)$ where $a,b,c,d$ are nonnegative integers? 

*what is the counting sequence for paths from $(0,0)$ to $(n,n)$ where $2n$ is the size of the path (number of steps) and n can vary over all nonnegative integers? Im not sure about this, but I think its $c(2n,n)$? I'm not so sure.
Any help would be very much appreciated, thank you!
 A: tldr version: yes your answers were all correct 

Trying to find how many different paths is equivalent to asking the question of how many different sequences of the multiset $\{x\cdot Right, y\cdot Up\}$ there are.
$$\begin{bmatrix} & & & \rightarrow&\rightarrow &\rightarrow &\bigcirc \\
 & & \rightarrow&\uparrow & & & \\
 & & \uparrow& & & & \\
\rightarrow & \rightarrow &\uparrow & & & & \end{bmatrix}$$
In my example pictured above, this sequence of movements corresponds to starting at the square $(0,0)$, moving right a total of 6 times and moving up a total of 3 times to the square at location $(6,3)$ and this particular sequence can be read as $R,R,U,U,R,U,R,R,R$
(note in my picture I'm using the center of the grid squares as the integer lattice, hence the ending circle.  Equivalent arguments can be made if you travel along the edges of squares instead of through the center of squares)
In the case you want to travel from $(0,0)$ to $(7,7)$, it corresponds to a total of 7 up movements, and 7 right movements.  Out of 14 total spaces in the sequence, a total of 7 of them will be taken up by ups (and all others will be rights), so it is an equivalent problem to "How many ways can you select 7 objects from a set of 14" and is $~_{14}C_7 = \binom{14}{7} = \frac{14!}{7!7!}$ as you correctly mentioned in your post.
In general, starting from $(x_1, y_1)$ and going to $(x_2, y_2)$ with $x_1\leq x_2$ and $y_1\leq y_2$ there will be a total of $((x_2-x_1) + (y_2 - y_1))$ motions total and $(x_2-x_1)$ of those will be rights (all others will be ups), for a total of $\binom{x_2-x_1+y_2-y_1}{x_2-x_1} = \frac{(x_2-x_1+y_2-y_1)!}{(x_2-x_1)!(y_2-y_1)!}$ number of ways.
Applying that to the case of starting from $(0,0)$ and going to $(n,n)$ it is indeed $\binom{2n}{n} = \frac{(2n!)}{n!n!}$

As an answer to your extra question in the comments, in the case that we have a starting point of $(0,0)$, an ending point of $(n,n)$, and wish to travel along a path of length $3n$, note first that to do so succesfully requires that $n$ be an even number.  Since to end at the goal requires an even number of moves, $3n$ will only be even if $n$ is even.  In other words, it is impossible if $n$ were odd.
Furthermore, by making a total of $3n$ number of moves, there will necessarily be some Lefts and/or Downs used as well.  (If we used only rights and ups, we would have passed the goal in either the right or up direction).  Also, there will be at least $n$ rights used and at least $n$ ups used (else, again, we couldn't reach the goal otherwise).
Let $x$ represent the number of extra rights used.  Then to end at the goal, there will have been a total of $n+x$ rights and a total of $x$ lefts.  This leaves a total of $n-2x$ moves total which will be split evenly between extra ups and downs (a total of $\frac{n-2x}{2}$ extra ups and $\frac{n-2x}{2}$ downs)
Since a sequence with $n+x$ rights is never equal to a sequence with $n+y$ rights if $x\neq y$, there is no overlap and we can count each case separately and add the results.  With $x$ extra rights it becomes the question of how many permutations exist of the multiset: $\{(n+x)\cdot Right, (n+(\frac{n-2x}{2}))\cdot Up, x\cdot Left, (\frac{n-2x}{2})\cdot Down\}$ for a total of $\dfrac{(3n)!}{(n+x)!(n+(\frac{n-2x}{2}))!x!(\frac{n-2x}{2})!}$ number of ways.
Letting $x$ range through all possible integer values, $0\leq x \leq \frac{n}{2}$ we get a final total of:
$$a_n = \begin{cases}\sum\limits_{x=0}^{\frac{n}{2}}\dfrac{(3n)!}{(n+x)!(n+(\frac{n-2x}{2}))!x!(\frac{n-2x}{2})!} ~~~ \text{if }n\text{ is even}\\ 0 \text{ otherwise}\end{cases}$$
I highly doubt that this will simplify further due to the heavy casework involved.
