MathRevolution wrote:
What is the number of the shortest routes from X to Y through Z?
A. 8
B. 12
C. 16
D. 18
E. 20
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!)]
--------------------------------------------------------------
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!) =
6So, 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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