What is the number of the shortest routes from X to Y through Z?

Nice question!

Let's first get from point X to point Z.
NOTE: we're trying to get to point Z via the SHORTEST route.
So, one possible route is: Up, Up, Right, Right (UURR)
Another possible route is: Right, Right, Up, Up (RRUU)
Another possible route is: Right, Up, Up, Right (RUUR)
etc

As you can see, the shortest route from point X to point Z will consist of 2 Ups (U's) and 2 Rights (R')
So, the question really becomes, "In how many different ways can we arrange 2 U's and 2 R's?"

-------ASIDE---------------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
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To get from point X to point Z, we must arrange 2 U's and 2 R's
There are 4 letters in total
There are 2 identical U's
There are 2 identical R's
So, the total number of possible arrangements = 4!/[(2!)(2!) = 6

So, there are 6 different routes from point X to point Z.

Once we arrive at point Z, we can see that there are only 2 possible ways to get to point Y:
1) Up then right
2) Right then up

So, for each of the 6 different routes to get to point Z, there are 2 ways to get to point Y
So, the TOTAL number of ways to get from point X to point Y (via point Z) = 6 x 2 = 12

Answer: B