Grid Problem in Maths related to number of ways to reach from one point to another Find the number of ways of reaching from one corner to diagonally opposite corner in 4 * 4 grid if movements are only along the lines and no point can be visited more than once . Also only allowed movements are vertically upwards, vertically downwards and right side movements. No left side movements are allowed.
I can solve this type of problem if only rightwards and vertically upwards movement are considered.
So, Please provide me solution to this problem. Any hint/solution will be greatly appreciated.
Thank You
 A: It's sometimes convenient to make a sketch of the problem. Let's consider a $4\times 4$ grid as given below.
                                   
We are looking for the number of paths from $A$ to $B$ where we are allowed to make up (U), down (D) and right (R) steps without ever passing a node more than once.

In order to go from $A$ to $B$ we have to take $4$ horizontal steps $R$. A choice of $4$ $R$ steps is given in the graphic by $a,b,c,d$.

*

*We observe there is one and only one path (marked in dashed green) from $A$ to $B$ which goes along $a\to b\to c\to d$.


*We are free to specify the height of $a,b,c,d$ and there are $5$ choices for each of these heights.
We conclude there are $\color{blue}{5^4=625}$ different valid paths from $A$ to $B$.

A: Assume you are going from bottom left corner to upper right corner.
To obtain $\frac{8!}{4!^2}=\binom{8}{4}$, you can think of words on the alphabet $\{U,R\}$ (for up and right) with 4 $U$s and 4 $R$s.
For your problem with three moves $\{U, R, D\}$, you will have 4 $R$s and 4 four more $U$s than $D$s, but you also have to make sure you stay in the grid ($U$ count $\ge D$ count at each step).  For a simpler approach, note that once you specify which "rung of the ladder" corresponds to each $R$ move, the path is completely determined.  There are $5$ choices (correction from @MarkusScheuer) for each $R$, and these choices are independent, so there are $5^4$ paths.
A: If vertical up and down movements are allowed, the answer is $5^4$.
Each time you move horizontally you can do so on any one of the horizontal links which form a vertical column.
As each point can be visited no more than once, there is one path between each horizontal link in one column, and each horizontal link in the next column.
Thus the answer is ${(\text{# horizontal links in a column})}^{(\text{# columns})} = 5^4$
