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krishnasty
Encountered another question and dont know how to proceed:

Every morning, Casey walks from her house to the bus stop. She always travels exactly nine blocks from her house to the bus, but she varies the route she takes every day. (One sample route is shown). How many days can Casey walk from her house to the bus stop without repeating the same route?

Similar questions to practice:
grockit-similar-to-og-quant-qustion-99962.html
pat-will-walk-from-intersection-a-to-intersection-b-along-a-68374.html
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The answer should be 126

9!/(4!*5!) or (9*8*7*6*5)/(4*3*2*1)
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Valerun
The answer should be 126

9!/(4!*5!) or (9*8*7*6*5)/(4*3*2*1)

Agree with 126 as answer

Namely 9C4 as also pointed out correctly by Quant Expert Bunuel

Answer: A

Cheers
J :)
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krishnasty
Encountered another question and dont know how to proceed:

Every morning, Casey walks from her house to the bus stop. She always travels exactly nine blocks from her house to the bus, but she varies the route she takes every day. (One sample route is shown). How many days can Casey walk from her house to the bus stop without repeating the same route?

In order to travel exactly nine blocks Casey should go 5 block down and 4 block left - DDDDDLLLL.

# of permutations of 9 letters DDDDDLLLL out of which there are 5 identical D's and 4 identical L's is \(\frac{9!}{5!4!}\).

Hope it's clear.


Why it cant be 6*5 (as in 6 lines to the left and 5 lines down?

or if we are choosing blocks than 5C1 * 4C1
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Bunuel
krishnasty
Encountered another question and dont know how to proceed:

Every morning, Casey walks from her house to the bus stop. She always travels exactly nine blocks from her house to the bus, but she varies the route she takes every day. (One sample route is shown). How many days can Casey walk from her house to the bus stop without repeating the same route?

In order to travel exactly nine blocks Casey should go 5 block down and 4 block left - DDDDDLLLL.

# of permutations of 9 letters DDDDDLLLL out of which there are 5 identical D's and 4 identical L's is \(\frac{9!}{5!4!}\).

Hope it's clear.


Why it cant be 6*5 (as in 6 lines to the left and 5 lines down?

or if we are choosing blocks than 5C1 * 4C1


sidoknowia,

Think of it like this:

There are a total of 9 turns that have to be made, 5D and 4L. In order to make find out the number of possible routes you have to think about how many different combinations of D turns you can make if there are 9 possible positions. Thinking of it like this leaves you with 9C5 = 9!/5!(9-5)! =126.

Or you cant think of it as how many combinations of L turns you cam make if there are 9 possible positions. This leaves you with 9C4 = 9!/4!(9-4)! = 126. Same answer either way.
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krishnasty
Encountered another question and dont know how to proceed:

Every morning, Casey walks from her house to the bus stop. She always travels exactly nine blocks from her house to the bus, but she varies the route she takes every day. (One sample route is shown). How many days can Casey walk from her house to the bus stop without repeating the same route?


to reach from start to end point one has to walk on 4 horizontal lines and 5 Vertical lines irrespective of the path chosen as only right and up movements are allowed.

Hence one of the possible route will be HHHHVVVVV

but another path could be VVHHVHHVV

i.e. all possible arrangements of these 9 letters HHHHVVVVV are different routes

the arrangements of these 9 letters can be represented by 9!/(5!*4!) which is same as 9C4 or 9C5

Alternatively we may say that out of 9 blank spaces we need to select only 4 places for H's which we can do in 9C4 ways and rest 5 places can be filled by V in just one way
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Madhavi1990
Agree with the solution. But, in the diagram it looks like its 8 blocks not 9 (5 on the right column and 3 on row)
Can anyone let me know what is my error here?

 
­In the diagram by block they mean the lines she needs to traverse.
So 4 such lines/blocks to the right and 5 such lines/blocks to the top as shown below.

Attachment:
Blocks.jpg
Blocks.jpg [ 10.31 KiB | Viewed 4651 times ]

Similar problem solved using 2 methods here



Hope it helps!­
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Madhavi1990
Agree with the solution. But, in the diagram it looks like its 8 blocks not 9 (5 on the right column and 3 on row)
Can anyone let me know what is my error here?
I have the same doubt!

Bunuel can you please help with this? Why are we taking 9! and not 8!?­
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ShreyaG7

Madhavi1990
Agree with the solution. But, in the diagram it looks like its 8 blocks not 9 (5 on the right column and 3 on row)
Can anyone let me know what is my error here?
I have the same doubt!

Bunuel can you please help with this? Why are we taking 9! and not 8!?­
­Here is an image that might help. Count the number of the line segments:


Attachment:
GMAT-Club-Forum-rrozjtpi.png
GMAT-Club-Forum-rrozjtpi.png [ 9.6 KiB | Viewed 1627 times ]
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Bunuel

ShreyaG7

Madhavi1990
Agree with the solution. But, in the diagram it looks like its 8 blocks not 9 (5 on the right column and 3 on row)
Can anyone let me know what is my error here?
I have the same doubt!

Bunuel can you please help with this? Why are we taking 9! and not 8!?­
­Here is an image that might help. Count the number of the line segments:

­Oh yes, that makes it very clear. Thank you Bunuel :)
Attachment:
GMAT-Club-Forum-9camqlue.png
GMAT-Club-Forum-9camqlue.png [ 9.6 KiB | Viewed 1618 times ]
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