The GMAT Math Pro logo is stuck on the highest ledge of a nine-story

Can you explain further? Thanks a ton!

If you observe the image in the problem, there are a total of 8 steps that the person must take to reach the bottom-most ledge. At each stage of this 8 step journey, the person can take only 2 decisions, whether to go right or to go left. However, since the left and right sides are constrained, i.e. a person can take a maximum of 4 left side jumps and 4 right side jumps, we must arrange these 4 lefts and 4 rights in a total of 8 steps.
Mathematically, 8!/(4!x 4!) (division by 4! twice to account for 4 lefts and 4 rights, similar to how we use this approach in problems on arranging words in a series)

I will try to give a good explanation here

So I will use the combinatorics method to demonstrate nick1816 method

To give you a clue on how we can count , I will assign L and R , meaning that Left and Right , if the guy chooses to go left , I will right L , if right i will write R

One path to reach the bottom is L(to 8th floor) , L(7th) , L(...) , L , R ,R ,R , R(bottom) (use the picture start from the top and then choose left for the first , second , third and fourth step , then you'll have only the right to go to the right bottom ( messing with your head a little).

so I will give you here a hint , all the paths to the bottom need to use 4 Left and 4 right (LRLLLRRR , LRLRLRLR....)
so this problem is a permutation one (Ah ha !!!!!!!!!!!)

since I have 8 choices and 4 repetitions in two sets .

number of roots = 8!/(4! * 4!)= 8C4 =70 Voila !!!!!!!!!!

E is the answer

(permutation rule , to arrange N letters for example , and you have A , B ,C letters repeted X, Y , Z times , then the number of words you can create from this list is N! / (X! * Y! * Z!)

You like the answer ? It's time for Kudos

This is going to be difficult to describe, but I’ll try my best.

If we count each “slot” as 1 unit, what we essentially have is a rhombus with 4 “slot-units” making up each side.

Shift the entire picture counterclockwise so that the stick man at the top is now in the bottom left corner.

The picture is then the “standard” picture in which the person can only travel East or North until he travels from the bottom-left corner to the upper-right corner of the figure.
However, instead of a perpendicular square, we have a bit of a “stretched” rhombus. The same logic still applies.

No matter which path you take, you will always be making 4 “Eastward” moves and 4 “Northward” moves.

E-E-E-E-N-N-N-N

Then just find the number of ways to arrange the 8 elements, of which 4 “E’s” are identical and 4 “N’s” are identical.

8! / (4! * 4!) = 70 ways

Clever way to alter the common question so as to make it more daunting. Great question!

Posted from my mobile device

Whatever path one takes, to reach the single bottom ledge, one has to take 4 left ledges and 4 right ledges.

One of the ways will be LLLLRRRR.

The total number of paths will be all the arrangements of the above path.

Thus, the total number of paths = 8!/4!*4! = 8*7*6*5/4*3*2*1 = 70 ways.

Thus, the correct option is B.