To go from X to Y, we need to go right twice (2 R's) and up three times (3 U's). Think of this question as how many ways you can order RRUUU. The answer is C. I go through the question in detail in the full solution below (taken from the GMATFix App).



-Patrick

kobel51 wrote:
Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the preceding map. How many routes form X to Y can Pat take that have the minimum possible length?

A) 6
B) 8
C) 10
D) 14
E) 16

Let E = traveling one block eastward and N = traveling one block northward.
To travel from X to Y by a route of minimum length, Pat must travel exactly 2 blocks eastward and exactly 3 blocks northward:
EENNN.
Any arrangement of the letters EENNN will yield a viable route.

The number of ways to arrange 5 DISTINCT elements = 5!.
But the elements above are not distinct: there are 2 identical E's and 3 identical N's.
When an arrangement includes IDENTICAL elements, we must DIVIDE by the number of ways to arrange the identical elements.
The reason is that the arrangement doesn't change when the identical elements swap positions, REDUCING the number of unique arrangements:

To illustrate:
The number of ways to arrange the letters in the word SPEED = 5!/2! = 60.
We divide by 2! to account for the 2 E's.
The number of ways to arrange the letters in the word RADAR = 5!/(2!*2!) = 30.
We divide by 2! to account for the 2 A's and by another 2! to account for the 2 R's.
The number of ways to arrange the letters in the word MISSISSIPPI = 11!/(4!*4!*2!).
We divide by 4! to account for the 4 S's, by another 4! to account for the 4 I's, and by 2! to account for the 2 P's.

In the problem at hand:
The number of ways to arrange EENNN = 5!/(2!*3!) = 10.

The correct answer is C.

Check here for a similar -- but trickier -- problem:

https://www.beatthegmat.com/different-routes-t93698.html