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19 Mar 2018, 10:58
Kudos
Bookmarks
Hi All,
There are a couple of ways that you can approach this question….
Since the answers are relatively small, there are at least 6 ways to get from X to Y, but no more than 16 ways to get from X to Y. In a pinch, you could draw pictures and physically find all of the possibilities.
If you're more interested in a "math" approach, you'll see that to get from X to Y you'll need to go 3 blocks "up" and 2 blocks "over" no matter how you get from X to Y.
Since you have to make 5 "moves" and 3 of them have to be "up", you have a Combination Formula situation….In other words…
5c3
5!/[3!2!] = 10
You COULD also say that to make 5 "moves" and 2 of them have to be "over", you could also use the combination formula in this way…
5c2
5!/[2!3!] = 10
It's the same answer because 5c3 is the same as 5c2.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
There are a couple of ways that you can approach this question….
Since the answers are relatively small, there are at least 6 ways to get from X to Y, but no more than 16 ways to get from X to Y. In a pinch, you could draw pictures and physically find all of the possibilities.
If you're more interested in a "math" approach, you'll see that to get from X to Y you'll need to go 3 blocks "up" and 2 blocks "over" no matter how you get from X to Y.
Since you have to make 5 "moves" and 3 of them have to be "up", you have a Combination Formula situation….In other words…
5c3
5!/[3!2!] = 10
You COULD also say that to make 5 "moves" and 2 of them have to be "over", you could also use the combination formula in this way…
5c2
5!/[2!3!] = 10
It's the same answer because 5c3 is the same as 5c2.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
27 Mar 2018, 10:01
Kudos
Bookmarks
Pat will walk 3 blocks north and 2 blocks east to get to Y.
This can be represented as NNNEE
Total ways = 5!
Because she walks North 3 times and East 2 times, that has to be accounted for as 3!*2! to capture the minimum routes.
Minimum possible paths are now:
5!/ (3!*2!) = 10
This can be represented as NNNEE
Total ways = 5!
Because she walks North 3 times and East 2 times, that has to be accounted for as 3!*2! to capture the minimum routes.
Minimum possible paths are now:
5!/ (3!*2!) = 10
11 Dec 2019, 08:29
Kudos
Bookmarks
haidzz
If we define Pat's path in a block-by-block manner, we can see that any route from X to Y will consist of 3 UPS and 2 RIGHTS.
So for example, if we let U represent walking one block UP, and let R represent walking one block RIGHT, one possible path is URURU.
Another possible path is UUURR
Another possible path is UURUR
So our question becomes, "In how many different ways can we arrange 3 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!)]
---------------------------------
Now let's apply the MISSISSIPPI rule to arranging 3 U's and 2 R's
There are 5 letters in total
There are 3 identical U's
There are 2 identical R's
So, the total number of possible arrangements = 5!/[(3!)(2!)] = 10
Answer: C
Cheers,
Brent
