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# In the figure above, X and Y represent locations in a district of a ce

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In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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21 Sep 2019, 02:23
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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 [ 9.64 KiB | Viewed 6958 times ]
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Posts: 8755
Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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21 Sep 2019, 04:33
5
6
gmatt1476 wrote:

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
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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04 Nov 2019, 22:04
5
1
We need to take 5 streets in East direction and 3 in North. It can be written for one combination as
EEEEENNN
So que is - in how many ways 5E's and 3N's can be arranged.
= 8!/(5!3!)=56
##### General Discussion
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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09 Oct 2019, 20:51
chetan2u wrote:
gmatt1476 wrote:

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

Thanks for your explanation. However, can you elaborate more why we just take 8C3 but not 8C3 + 8C5 (as i think that there are 2 possibilities, go from north 8C3 and go from east 8C5), please show me where im wrong. Thank you so much
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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13 Oct 2019, 16:23
chetan2u wrote:
gmatt1476 wrote:

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

chetan2u : I am always confused about when to multiply probability and when to add probability. What is the logic behind this in probability problems
Math Expert
Joined: 02 Aug 2009
Posts: 8755
Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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13 Oct 2019, 18:24
StudiosTom wrote:
chetan2u wrote:
gmatt1476 wrote:

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

chetan2u : I am always confused about when to multiply probability and when to add probability. What is the logic behind this in probability problems

Hi,
It will depend on what the question is asking..
So, in very simple terms, I would say

Multiply..
Say you have 10 different things and we are looking for probability to pick a particular item on three different times when we pick one item..
If you see the probability should be DECREASING.. Picking that item is 1/10 in each pick so it becomes 1/10*1/10*1/10
A coin- getting tails on two continuous throws or tails in first throw and heads in second throw---- 1/2*1/2
So, generally we are looking at multiple events and probability of a particular/different item each time
That is we have AND .. x and y happening simultaneously

Say you have 10 different things and we are looking for probability to pick any of two different item when we pick one item..
If you see the probability should be INCREASING, as now we have options... Picking each item is 1/10 in each pick so picking any of the two becomes 1/10+1/10
A coin- getting tails or heads on a throw ---- 1/2+1/2=1, as it is sure that we will get heads or tails in that throw
So, generally we are looking at a single events and probability of more than one item.
That is we have OR.. option of either of x or y happening
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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14 Oct 2019, 08:04
4
Hungluu92vn wrote:
chetan2u wrote:
gmatt1476 wrote:

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

Thanks for your explanation. However, can you elaborate more why we just take 8C3 but not 8C3 + 8C5 (as i think that there are 2 possibilities, go from north 8C3 and go from east 8C5), please show me where im wrong. Thank you so much

chetan2u.

is it basically we are moving 5 units right & 3 units left whatever route we take to reach from X to Y, this means this is basically arrangement of 8 objects among which 5 are identical of one type & 3 are of others.?

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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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03 Dec 2019, 13:20
1
J2S2019 wrote:
Hungluu92vn wrote:
gmatt1476 wrote:

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

Thanks for your explanation. However, can you elaborate more why we just take 8C3 but not 8C3 + 8C5 (as i think that there are 2 possibilities, go from north 8C3 and go from east 8C5), please show me where im wrong. Thank you so much

chetan2u.

is it basically we are moving 5 units right & 3 units left whatever route we take to reach from X to Y, this means this is basically arrangement of 8 objects among which 5 are identical of one type & 3 are of others.?

You are correct. Since there are 8 options in total, but 5 right and 3 up are the exact same, we use the 8!/3!x5!

Keep it up!
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Joined: 19 Sep 2019
Posts: 32
Location: Austria
Concentration: General Management, Organizational Behavior
Schools: HSG SIM "22 (A)
GMAT 1: 730 Q48 V42
Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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03 Dec 2019, 13:24
Hungluu92vn wrote:
chetan2u wrote:
gmatt1476 wrote:

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

Thanks for your explanation. However, can you elaborate more why we just take 8C3 but not 8C3 + 8C5 (as i think that there are 2 possibilities, go from north 8C3 and go from east 8C5), please show me where im wrong. Thank you so much

Remember the formula for how to arrange a word like MISSISSIPPI?

Well, if you use 11!, you will have a couple of times the exact same word within that combination, since there are 4S for example, the formula pretends that the 4 S are different.

Now in the example, you can go up 3 times (UpUpUp=UUU) and right 5 times (RightRightRightRightRight=RRRRR). So it is basically the combo of UUURRRRR and as in MISSISSIPPI, you wanna avoid counting double. Thus you divide 8! by the number of times U and R repeats (as !)

That is why it is:

8!/5!x3!
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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07 Dec 2019, 16:22
2
1
Hi All,

We're told that X and Y represent locations in a district of a certain city where the streets form a rectangular grid. We're asked to travel ONLY North OR EAST along the streets from X to Y. We're asked for the number of different paths that are possible. This question is a variation on the Combination Formula.

To start, since we can ONLY travel North OR East, we will eventually have to travel 3 streets "up" and 5 streets "to the right" (in some order). That ultimately means that we will have 8 total "moves." Sometimes we'll be able to choose to go North or East, but sometimes we'll have no choice (we'll only be able to North or only be able to go East). The order of the "up" moves and "to the right" moves does NOT matter though, as long as we get from X to Y. Since the order does not matter, that's a hint to use the Combination Formula:

N!/(K!)(N-K)!

Here, N = 8 (since there are 8 total moves) and we can make the K equal either 3 or 5 (for the 3 "up" moves or 5 "to the right" moves). Either calculation will end in the same value...

8!/(3!)(5!) or 8!/(5!)(3!)

This will give us (8)(7)(6)/(3)(2)(1) = (8)(7)/1 = 56 possible ways to get from X to Y.

GMAT assassins aren't born, they're made,
Rich
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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13 Dec 2019, 13:32
gmatt1476 wrote:

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

Since there are three blocks along the north to south direction and five blocks along the east to west direction, a path from X to Y will consist of 8 one-block segments, of which 3 segments go north and 5 segments go east. Therefore, the number of different paths is:

8! / (3! x 5!) = (8 x 7 x 6 x 5!) / (3 x 2 x 5!) = 8 x 7 = 56

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In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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03 Jan 2020, 15:18
MBA HOUSE KEY CONCEPT: Combinatorics Permutation with repetition.

8 steps = 5 to east + 3 to north

Permutation of 8 with 5 and 3 repetitions = 8! / (5!)(3!) = 56

E
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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13 Jun 2020, 04:29
i am still not able to visualize what results 5*3 implies.You need to choose East ways = 5 times and north = 3 times, then total ways = 5*3= 15?
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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13 Jun 2020, 13:05
1
Hi SkR1,

You might find it helpful to deal with this prompt in small 'pieces'

First, we're told that we can travel ONLY North OR East along the streets from X to Y. We're asked for the number of different paths that are possible to get from X to Y.

Second, we will eventually have to travel 3 streets "North" and 5 streets "East" (in some order). That ultimately means that we will have 8 total "moves." For example, the 8 moves could be:

NNNEEEEE
EEEEENNN
NENENEEE
Etc.

The order of the "North" moves and "East" can be in any order though, as long as we get from X to Y. Since the order does NOT matter, that's a hint to use the Combination Formula:

N!/(K!)(N-K)!

Here, N = 8 (since there are 8 total moves) and we can make the K equal either 3 or 5 (for the 3 "up" moves or 5 "to the right" moves). Either calculation will end in the SAME value...

8!/(3!)(5!) or 8!/(5!)(3!)

Solving that calculation gives us all of the possible ways to get from X to Y.

GMAT assassins aren't born, they're made,
Rich
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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13 Jun 2020, 23:01
EMPOWERgmatRichC wrote:
Hi SkR1,

You might find it helpful to deal with this prompt in small 'pieces'

First, we're told that we can travel ONLY North OR East along the streets from X to Y. We're asked for the number of different paths that are possible to get from X to Y.

Second, we will eventually have to travel 3 streets "North" and 5 streets "East" (in some order). That ultimately means that we will have 8 total "moves." For example, the 8 moves could be:

NNNEEEEE
EEEEENNN
NENENEEE
Etc.

The order of the "North" moves and "East" can be in any order though, as long as we get from X to Y. Since the order does NOT matter, that's a hint to use the Combination Formula:

N!/(K!)(N-K)!

Here, N = 8 (since there are 8 total moves) and we can make the K equal either 3 or 5 (for the 3 "up" moves or 5 "to the right" moves). Either calculation will end in the SAME value...

8!/(3!)(5!) or 8!/(5!)(3!)

Solving that calculation gives us all of the possible ways to get from X to Y.

GMAT assassins aren't born, they're made,
Rich

5*3 means just go from point A to ( 5 ways) X to (3 ways) B - which is not the case in this question
i got your point, thanks for clarification
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Re: In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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29 Jun 2020, 08:00
itsSKR wrote:
i am still not able to visualize what results 5*3 implies.You need to choose East ways = 5 times and north = 3 times, then total ways = 5*3= 15?
Mbahousegmat ScottTargetTestPrep EMPOWERgmatRichC

Response:

One way the answer could simply be found by calculating 5 * 3 = 15 is if you had to first travel east to the point directly below Y and then travel north, and there were 5 different paths going to the point on the east and there were 3 different paths going from that point to Y. It is nothing like the situation we have in this question.

Instead, what we have in this question is that you have the option to either travel north or east at each step and the total number of east steps should be 5 and the total number of north steps should be 3. For instance, you could first go all the way to the east to the point directly below Y and then go all the way to the north to the point Y. This path can be denoted as EEEEENNN (five east steps followed by three north steps). Alternatively, you could first travel all the way to the north and then travel east. That path would be NNNEEEEE. There are also other paths, such as NENENEEE (take alternating steps between north and east until you hit the top line, then keep traveling east) or EEENNNEE (take three steps to the east, three steps to the north and another two steps to the east).

So, you should observe that as long as there are three N’s and five E’s in the word you write, you get a path from X to Y. Furthermore, any path from X to Y can be described with a word containing three N’s and five E’s. Thus, the total number of paths is equal to the number of ways you can arrange the word EEEEENNN. Once you make this observation, the answer can easily be found by applying the permutations of indistinguishable objects formula, which is 7!/(5!*3!).
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In the figure above, X and Y represent locations in a district of a ce  [#permalink]

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07 Jul 2020, 00:43
We can go only above and to the right

Consider 1 step to the right as 1H (horizontal) and 1 step above as 1V (vertical)

So in total we have 3 verticals i.e. 3 V's and 5 horizontals i.e. 5 H's

Choose any route, you would have to require all 3 V's and all 5 H's

Thus you can visualize this as a question of arranging letters in which the order doesnt matter, there are 8 such letters (VVVHHHHH or HHHHHVVV or VVHHHHHV etc)
= 8!

3 V's are repeated and so are 5 H's
Hence divide 8! by 3!*5!

Thus final answer = $$\frac{8!}{5!*3!}$$ i.e. 56
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In the figure above, X and Y represent locations in a district of a ce   [#permalink] 07 Jul 2020, 00:43