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gmatt1476

In the figure above, X and Y represent locations in a district of a certain city where the streets form a rectangular grid. In traveling only north or east along the streets from X to Y, how many different paths are possible?

A. 720
B. 512
C. 336
D. 256
E. 56


PS61551.01

Attachment:
2019-09-21_1421.png

If we define paths from X to Y in a block-by-block manner, we can see that any route from X to Y will consist of 3 NORTHS and 5 EASTS.

For example, this particular path....

... can be defined as ENENEEEN (where E stands for 1 block east, and N stands for 1 block north.

Similarly, this path...

... can be defined as NNENEEEE

So our question becomes, "In how many different ways can we arrange 3 N's and 5 E'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 N's and 5 E's
There are 8 letters in total
There are 3 identical N's
There are 5 identical E's
So, the total number of possible arrangements = 8!/[(3!)(5!)] = 56

Answer: E

Cheers,
Brent
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This is a classic combinatorics question of permutation with repetition where you put the total possibilities factorial in the numerator and the repetitions factorial in the denominator.
You always walk 5 steps to the right and 3 to the top whatever the possibility that you choose.

Permutation of 8 steps with 3 and 5 repetitions.

8!/3!5!= 56

E
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This is a really excellent question.

Don't let the diagram frighten you

This question is all about out of the "grid" thinking. ;)

There can be n number of ways to reach point B from A, but all the paths must have 5 east and 3 north

EEEEENNN

So how many ways can be there? As many arrangements are there for the above

8! / 5!*3!
= 56 ways

I hope it helps! :)
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Do questions like these (questions related to direction and counting questions that need us to count the number of ways of reaching one point to another) which are in the GMAT Official Advanced questions book(2020) show up in the newer GMAT Focus Edition?
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SPM1234
Do questions like these (questions related to direction and counting questions that need us to count the number of ways of reaching one point to another) which are in the GMAT Official Advanced questions book(2020) show up in the newer GMAT Focus Edition?
­The short answer would be - yes.
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gmatt1476

In the figure above, X and Y represent locations in a district of a certain city where the streets form a rectangular grid. In traveling only north or east along the streets from X to Y, how many different paths are possible?

A. 720
B. 512
C. 336
D. 256
E. 56


PS61551.01

Attachment:
2019-09-21_1421.png


If you look at the streets, you have to move 5 streets in the east and 3 streets to the north..

Thus, the total streets are 5+3 and the 3 N or 5 E can be chosen out of these in any order..

So, total ways = 8C3=\(\frac{8!}{5!3!}=\frac{8*7*6}{3*2}=8*7=56\). Hence 56 ways.

E
Hi, where does the 3 comes in from in 8C3 ?
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Deconstructing the Question

To get from X to Y, we can move only east or north.

So the problem becomes: how many ways can we arrange the required east and north moves?

Step-by-step

Count the grid carefully.

There are \(5\) moves east and \(3\) moves north.

So each path has a total of

\(5 + 3 = 8\)

moves.

Now choose where the \(3\) north moves go among the \(8\) total moves:

\(\binom{8}{3}\)

Compute:

\(\binom{8}{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56\)

Answer E
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