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
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==> The number of cases of x heading to z is 4!/2!2!=6. The number of cases of Z heading to y is 2, which makes 6*2=12.

Hence, the answer is B.
Answer: B
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No matter what, you have to take 2 horizontal steps and 2 vertical steps to reach point "Z" from "X".

let's Horizontal step=H
vertical step= V

"HHVV" the no. of ways to arrange them= \(\frac{4!}{2!*2!} = 6\) ways to reach Z

the no. of ways to reach Y from Z only \(2 ways\)

now, multiply the results= \(6*2= 12\) ways total

Answer B

If I see a question like this again, how do I tell it is a combination question?

=> Best Approach: Regard the different parts of the shortest route as repeated letters in a word
=> Start at X, End at Y, Stop at Z

Two parts: Shortest route (X to Z) & Shortest route (Z to Y)

Modify the question: X to Z
Stretches of vertical roads/ steps = ‘a’
Stretches of horizontal roads/ steps = ‘b’
=> List all stretches of roads: b,b,a,a

=> Total number of roads: 4 (2 b^′ s & 2 a^′ s)

=> "∴" Formula: 4!/(2! 2! ) = 6

=> Modify the question: Z to Y
Stretches of vertical roads/steps = ‘p’
Stretches of horizontal roads/steps = ‘q’

=> List all stretches of roads: p,q

=> Total number of roads: 2 (1 p^′ s & 1 q^′ s)

=> " ∴" Formula: 2! = 2

=> Shortest routes: 6 routes (X~Z) and 2 routes (Z~Y)

=> " ∴" Total possible shortest routes: 6 * 2 = 12

Thanks
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