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Re: Pat will walk from intersection X to intersection Y along a route that
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19 Mar 2018, 09:58
1
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.
Re: Pat will walk from intersection X to intersection Y along a route that
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11 Dec 2019, 07:29
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haidzz wrote:
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?
If we define Pat's path in a block-by-block manner, we can see that any route from X to Y will consist of 3 UPS and 2 RIGHTS. So for example, if we let U represent walking one block UP, and let R represent walking one block RIGHT, one possible path is URURU. Another possible path is UUURR Another possible path is UURUR
So our question becomes, "In how many different ways can we arrange 3 U's and 2 R'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 U's and 2 R's There are 5 letters in total There are 3 identical U's There are 2 identical R's So, the total number of possible arrangements = 5!/[(3!)(2!)] = 10
Answer: C
Cheers, Brent
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Re: Pat will walk from intersection X to intersection Y along a route that
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17 Dec 2019, 19:52
haidzz wrote:
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?
Let V denote a step in the vertical direction and H denote a step in the horizontal direction. For instance, V-V-V-H-H denotes the path of walking along Avenue A until the intersection of 4th street and walking along 4th street until the point Y. Similarly, V-H-V-H-V denotes the path of walking along Avenue A, then walking along 2st street, then walking along Avenue B, then walking along 3rd street and, finally, walking along Avenue C to reach point Y.
We notice that a shortest path between point X and Y must include three V’s and two H’s. Further, any arrangement of three V’s and two H’s (i.e., any arrangement of the letters V-V-V-H-H) gives us a shortest path between X and Y. Using the permutations with indistinguishable objects formula, we see that there are 5! / (3!*2!) = (5 x 4)/2 = 10 such arrangements. Thus, there are 10 shortest paths between points X and Y.
Pat will walk from intersection X to intersection Y along a route that
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01 Feb 2020, 11:14
Bunuel wrote:
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.
Answer: C.
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.
Re: Pat will walk from intersection X to intersection Y along a route that
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01 Feb 2020, 11:29
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Top Contributor
Kritisood wrote:
Bunuel wrote:
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.
Answer: C.
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.
Re: Pat will walk from intersection X to intersection Y along a route that
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06 Feb 2020, 11:03
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Kritisood wrote:
Bunuel wrote:
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.
Answer: C.
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.
Silly doubt but why do we use permutations/arrangements here if order doesnt matter?
If he goes URURU or UUURR, the order doesnt matter right?
The question is asking for the number of routes from X to Y which has the minimum number of lengths, i.e. 5 step routes. The routes URURU and UUURR are considered to be different routes and thus, the order matters in this question. That's why we are using permutations (with indistinguishable objects).
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Re: Pat will walk from intersection X to intersection Y along a route that
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14 Mar 2020, 17:33
Just so I can understand this concept better and wording of this question, would the solution be different for maximum possible length instead of the minimum possible length?
Re: Pat will walk from intersection X to intersection Y along a route that
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19 Apr 2020, 19:40
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?
Re: Pat will walk from intersection X to intersection Y along a route that
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20 Apr 2020, 00:44
Asad wrote:
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?
in case of maximum length the answer is be Infinite
Cause the traveller can continue to remain in a loop infinitely.
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Re: Pat will walk from intersection X to intersection Y along a route that
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21 Apr 2020, 04:16
Asad wrote:
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?
Also, is my counting right for the maximum possible length?
I appreciate your help. Thanks__
Hello Asad,
Because you were counting the number of routes with the minimum possible length is why you considered that you had to travel only in the right and upward direction; your objective was to minimize the length.
As GMATinsight has already mentioned, the maximum possible length of the route will be infinite. Remember, you are looking for routes with maximum LENGTH. Length can be maximized by not reaching the destination and continuing to go around in circles in the grid.
And that’s exactly why the question is about finding the routes with the minimum possible length.