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# Pat will walk from intersection A to intersection B along a

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Re: Pat will walk from intersection A to intersection B along a  [#permalink]

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19 Mar 2018, 09:58
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.

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Re: Pat will walk from intersection A to intersection B along a  [#permalink]

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27 Mar 2018, 09:01
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
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Pat will walk from intersection A to intersection B along a  [#permalink]

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29 Apr 2018, 06:13
Quote:

Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length?
A) 6
B) 8
C) 10
D) 14
E) 16

In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.

So combination of UUURR: # of permutations of 5 letters out of which there are 3 identical U's and 2 identical R's is 5!/3!2!=10.

If there were 5 streets and 4 avenues then the answer would be combination of UUUURRR: # of permutations of 7 letters out of which there are 4 identical U's and 3 identical R's is 7!/4!3!=35.

Similar questions:
http://gmatclub.com/forum/grockit-simil ... 99962.html
http://gmatclub.com/forum/casey-and-the-bus-104236.html

Bunuel why do you assume that nn order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.

going three times up and two times left means we have to go throgh five lines of the same length right ? if so, then all routes are of equal length should we go at girst left than right , or left...we still cross the same number of of lines of identical length (5 lines and all are of the same length )

please proof that i am wrong
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Joined: 01 Jan 2018
Posts: 1
Re: Pat will walk from intersection A to intersection B along a  [#permalink]

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03 Nov 2018, 11:26
The question I'm about to ask probably reveals my competency level when it comes to combinatorics & probability....workin on it!

I haven't sifted through the responses on this thread yet but what would happen if we were asked, "How may routes from X to Y can Pat take that have the MAXIMUM possible length?
Re: Pat will walk from intersection A to intersection B along a &nbs [#permalink] 03 Nov 2018, 11:26

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