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Every morning, Casey walks from her house to the bus stop [#permalink]

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04 Nov 2010, 07:05

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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?

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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 \(\frac{9!}{5!4!}\).

Re: Every morning, Casey walks from her house to the bus stop [#permalink]

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24 Sep 2013, 22:43

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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?

Re: Every morning, Casey walks from her house to the bus stop [#permalink]

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30 Jun 2015, 06:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Every morning, Casey walks from her house to the bus stop [#permalink]

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08 Jul 2016, 00:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Every morning, Casey walks from her house to the bus stop [#permalink]

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07 Sep 2016, 16:29

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Bunuel wrote:

krishnasty wrote:

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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Re: Every morning, Casey walks from her house to the bus stop [#permalink]

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21 Sep 2016, 09:36

sidoknowia wrote:

Bunuel wrote:

krishnasty wrote:

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.

gmatclubot

Re: Every morning, Casey walks from her house to the bus stop
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21 Sep 2016, 09:36

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