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655-705 Level|   Combinations|                        
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haidzz

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. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen

PS45461.01

Attachment:
2019-09-21_1430.png

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.

Answer: C
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MBA HOUSE KEY CONCEPT: Combinatorics Permutation with repetition

5 steps = 2 to the right and 3 upward

Permutation of 5 with 2 and 3 repetitions = 5! / (2!)(3!) = 10

C :shocked
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Bunuel
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.


Hi Bunuel GMATPrepNow ScottTargetTestPrep

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?
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Bunuel
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.


Hi Bunuel GMATPrepNow ScottTargetTestPrep

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?

URURU and UUURR are considered different paths.
So, order does matter.
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Bunuel
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.


Hi Bunuel GMATPrepNow ScottTargetTestPrep

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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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?
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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. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen
Hello experts,
EMPOWERgmatRichC, VeritasKarishma, Bunuel, chetan2u, IanStewart, ArvindCrackVerbal, AaronPond, GMATinsight, JeffTargetTestPrep, ccooley
What if the word ''minimum'' has be changed with ''maximum''?
Which one will be correct counting?
1/
Total --11
Right--6
Up--5

Or,

2/
Total--11
Right--4
Up--4
Back--2
Down--1

Also, is my counting right for the maximum possible length?

I appreciate your help.
Thanks__
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Asad
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. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen
Hello experts,
EMPOWERgmatRichC, VeritasKarishma, Bunuel, chetan2u, IanStewart, ArvindCrackVerbal, AaronPond, GMATinsight, JeffTargetTestPrep, ccooley
What if the word ''minimum'' has be changed with ''maximum''?
Which one will be correct counting?
1/
Total --11
Right--6
Up--5

Or,

2/
Total--11
Right--4
Up--4
Back--2
Down--1

Also, is my counting right for the maximum possible length?

I appreciate your help.
Thanks__

Asad

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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Asad
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. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen
Hello experts,
EMPOWERgmatRichC, VeritasKarishma, Bunuel, chetan2u, IanStewart, ArvindCrackVerbal, AaronPond, GMATinsight, JeffTargetTestPrep, ccooley
What if the word ''minimum'' has be changed with ''maximum''?
Which one will be correct counting?
1/
Total --11
Right--6
Up--5

Or,

2/
Total--11
Right--4
Up--4
Back--2
Down--1

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.

Hope that helps!
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Similar problem solved using 2 methods here



Hope it helps!
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TallJTinChina
Bringing up an old topic because it is giving me headaches. :shock: :shock: :shock:

I understand breaking this down into 5 steps (2 Rights, 3 Ups)

So right = R
And Up = U

So we need to see how many permutations include 2 Rights, and 3 Ups

So I made a list

Start to the right
RRUUU
RURUU
RUURU
RUUUR

Start with by moving up
URRUU
URURU
URUUR
UURRU
UURUR
UUURR

Obviously making a list is not the quickest way to solve this problem.

How do I get to it this ----> 5!/2!3! = 10 ????????

It seems to be me are picking 5 out of 5? So we would have 5!/5!0! which equals 1

you have 5 steps to get to Y. either U or R. _ _ _ _ _. since you need to pick the shortest way, you know that already 3 of the underlined spots will be filled with U. you can pick 2 spots out of 5 for the Rs. Therefore to pick 2 out of 5 for the Rs you do a 5C2 = 5!/2!3!(and therefore you mandatorily allocate the rest to the U's)
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This is a classic combinatorics question of permutation with repetition where you put the total possibilities factorial in the numerator and the repetitions factorial in the denominator.
You always walk 2 steps to the right and 3 to the top whatever the possibility that you choose.

Permutation of 5 steps with 2 and 3 repetitions.

5!/2!3!= 10

C
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Can someone help me understand what’s the mistake behind 2^6 ?

There are 6 points where you have 2 directions and the others points you only have 1 possibility
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JJDa
Can someone help me understand what’s the mistake behind 2^6 ?

There are 6 points where you have 2 directions and the others points you only have 1 possibility

Hi JJDa,

Since you have not explained your thinking, I'm going to assume that you are counting each 'intersection' that you would take when traveling the shortest length from X to Y (including the intersections at X and Y); that would total 6 intersections. However, the minimal length is only 5 'segments' - meaning that since you're STARTING at X, you should NOT count that intersection. In addition, at each step in the journey, you will NOT necessarily have 2 possible choices. Draw any the shortest possible individual routes from X to Y and you'll find that at some point, the remaining step(s) would leave you with just 1 option.

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

Contact Rich at: [email protected]
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