Counting Paths using combinations and permutations Suppose I am given a grid of 2x2 , with A as starting point and B as the destination.  We are allowed to move either vertically up by 1 step or towards right by 1 step.  Through visual inspection, it is easy to find 6 ways to reach from A to B

But I am facing a difficulty in doing it by Permutations/Combinations
My thought:
I figured out that we need two right turns (R1 , R2) and two up steps (U1, U2)
where R1 = first right turn , R2 = second right turn
U1 = first up movement , U2 = second up movement
Now each possible way will have some combination of these steps, I am not sure how to go forward from here
After referring to the solution it mentioned the ways to be 4!/(2! * 2!) OR 4c2 * 2c2
so I started making cases for 4c2 * 2c2

*

*R1 R2   U1 U2

*R1 U1   R2 U2

*R1 U2   R2 U1

*R2 U1   R1 U2

*R2 U2   R1 U1

*U1 U2   R1 R2

So aren't cases 2,3,4,5 same thing only ?  and how will the case 3 for example be possible R1 U2 , how can we get U2 before than U1 ?
 A: We need to take a total of 2 vertically up steps and 2 steps towards right (i.e., we have 2 different types of steps), that we can take in any order: we can do it in $\dfrac{(2+2)!}{2!2!}=6$ ways.
Generalizing, let's denote a single up step by a red object and right step by a blue object. Then number of ways to construct a path of total m+n length which contains m up and n right steps is exactly same as number of ways to sort / permute m red and n blue objects, which is  $\dfrac{(m+n)!}{m!n!}$. Here, we have $m=n=2$.
A: Solution 1:
Note that to reach point B from point A, you need to move 2 points to the right and 2 points upwards, in total, you'd have to perform 4 moves. This is equivalent to arrange the following 4 elements:
$$
\{\text{upwards}, \text{upwards}, \text{right}, \text{right}\}
$$
For example, you can have the outcome $(\text{right}, \text{upwards}, \text{upwards}, \text{right})$, this outcome means that in the first move, you take 1 step to the right, for the second move you take 1 step upwards and so on. However, when you arrange those 4 elements, you are supposing that the 4 elements are distinct, but right and upwards are repeated. So, we need to divide the total number of arrangements of $\{\text{upwards}, \text{upwards}, \text{right}, \text{right}\}$ (which is 4!) by the product of all the arrangements of the repeated options, that is,
$$
\{\text{right}, \text{right}\}\quad \text{can be arranged in 2! ways}\\
\{\text{upwards}, \text{upwards}\}\quad \text{can be arranged in 2! ways}
$$
Then, the number of paths from A to B, supposing that you are allowed to move only to the right or upwards is:
$$
\frac{4!}{2!*2!}=6
$$
Solution 2: For this solution, let's use combinations instead of permutations. You'd have to perform 4 moves; 2 to the right and 2 upwards. So, suppose that you have 4 empty spaces (each space corresponds to 1 move) and you want to assign to each space one move either to the right or upwards. You can start selecting the spaces where you're going to move to the right (you'll get the same answer if you first select where you move vertically up), note that each choice uniquely determines one path:
$$
{{4}\choose{2}}*{{2}\choose{2}}=6*1=6
$$
