The following diagram shows the road map of city. The lines through th

Question

The following diagram shows the road map of city. The lines through the city indicates roads, but there is no road pass through the park. All the roads are either parallel or perpendicular to each other. Peter wants to go from X to Y by travelling the minimum possible distance. In how may ways can he make the journey?

A. 55
B. 100
C. 110
D. 166
E. 210

A. 55
B. 100
C. 110
D. 166
E. 210

MahmoudFawzy · Accepted Answer

I don't guess it is a GMAT question, but I will give it a try just for fun.

I try to calculate the number of ways for each point, one by one.
Each point is the sum of the two (or one) preceding points (i.e only points below or leftside in order to maintain least distance)
So at some points around the park, the number of ways is the same because there is only one approach to this point.

I am trying to explain it by words as I can, but the drawing below can better illustrate the number of ways for each point in the way to the final destination "Y"
and it will be 110.

Kinshook · Answer

Given: The following diagram shows the road map of city. The lines through the city indicates roads, but there is no road pass through the park. All the roads are either parallel or perpendicular to each other. Peter wants to go from X to Y by travelling the minimum possible distance.

Asked: In how may ways can he make the journey?

The problem is similar to choosing LLLLRRRRRR (4Ls & 6Rs) since travelling 4 times in left and 6 times in right direction, he will reach the destination with minimum possible distance

No of ways (without considering park) = 10C4 = 210

Because of existence of the park, some movements are not possible, (2L , 3R) combination is not possible

No of ways = 5C2 = 10 to reach the centre of the park

Further no of ways to reach destination from centre of the park = 5C2 = 10

Total no of ways to move through centre of the park = 5C2 * 5C2 = 10*10 = 100

Total no of ways to reach the destination without entering the park = 210 -100 = 110

IMO C

nick1816 · Answer

MahmoudFawzy Similar Official Question- https://gmatclub.com/forum/in-the-figur ... 05848.html I don't guess it is a GMAT question , but I will give it a try just for fun.

ManjariMishra · Answer

chetan2u ,  Bunuel ,  VeritasKarishma  Could you please explain how to omit park in this question

nick1816 · Answer

Total number of ways Peter can make journey from X to Y without any restrictions = 10!/4!*6!=210

Peter can't go through point Z; hence, eliminate all the routes that includes point Z.

Routes from X to Z= 5!/2!*3!= 10

Routes from Z to Y= 5!/2!*3!=10

Total number of routes that include point Z= 10*10=100

Number of ways he can make journey= 210-100=110

Fdambro294 · Answer

for some reason the diagram was not showing up on my computer, but it is probably my beat up laptop that I've been using to study......

The way I like to think of these problems is similar to the "Stars and Bars" Method used to find the No. of Distributions of Similar Items to Distinct Groups.

Ignoring the Park in the middle of the Map and FIRST assuming that we have a perfectly formed grid map with NO Park:

(1st)since he wants to travel the MINIMUM Distance to get to Y, he will only want to keep moving towards his objective. Therefore, at each Decision Point he will only want to move North (N) or East (E) ----- and never backwards (West) or downwards (South)

(2nd)again, assuming that the Park is filled in and we have a perfect grid, if we were to find every single possible MIN route he could take from X to Y, we would immediately notice that every Route would include:

6 moves East (E)

and

3 moves North (N)

he could go:
E-E-E-E-E-E-N-N-N-N
or
E-E-E-E-E-N-N-N-E-N
etc.

So if were to label pieces of paper with these Letters and scatter them on the floor and figure out how many different ways we could Shuffle the N's in and around all of the E's, we can find all the Unique and Distinct Routes that are possible.

We can calculate the above by using the formula for Arranging Identical Elements or using the "Stars and Bars" Method (pretending the N's are the Identical Partitions)

(6 + 4)! / (6! * 4!) = (10)! / (6! * 4!) = 210 ways

(2nd) However, because we can not travel through the Park, we need to Eliminate every possible Route that includes passage through the Park. In effect, we have over-counted the Actual Routes that are allowed under the constraints of the problem. We need to eliminate any Route that includes passage through the park.

Placing a dot immediately in the center of the park (Labeled Z) and filling in the grid with 3 more line segments that connect to Point Z, the routes that are NOT Allowed and that we need to Eliminate include all the different "shufflings" of:

-1st- the Unique Routes to get from X-to-Z

AND

-2nd- the Unique Routes to get from Z-to-Y

Finding all the Different Arrangements of these 2 Legs will give us all the Routes of the ENTIRE journey from X to Y that include passage through the Park. We can then eliminate all of these routes from the 210 over-count above as they are NOT allowed.

Using the Same Logic to find these routes that are NOT allowed:

-1st- Leg 1: Every single Minimum Route from X-to-Z will include:

3 Moves East (E)
and
2 Moves North (N)

Shuffling around the Different N's and E's again, we will get routes like:

E-E-E-N-N
or
E-E-N-E-N
etc.

the Number of different ways we can shuffle these Identical N's and E's around is equal to =

(5!) / (3!*2!) = 10 Distinct Routes for Leg 1

AND

-2nd- Leg 2: since we need to complete the entire journey to Point Y, we must find the Same No. of Unique Routes from Point Z to Point Y.

Again, there will always be 3 Moves East (E) and 2 Moves North (N)

(5!) / (3!*2!) = 10 Distinct Routes for Leg 2

Finally, to find the Total No. of Distinct Routes for the ENTIRE Journey from Point X to Point Y:

we have 10 Distinct Routes for Leg 1

and

these 10 Distinct Routes for Leg 1 can be shuffled and arranged with the 10 Distinct Routes for Legs 2 to find ------>

10*10 = 100 different Routes through the Park that we should not have counted originally.

(210 Original Over-Count) - (100 Routes NOT allowed) = 110 Possible Routes

Very difficult variation on a problem-type that has shown up in the Official Guide a few times......Nice 750+ type problem!

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