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# In certain city, a car is parking at A destination, it will go to B de

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Math Expert
Joined: 02 Sep 2009
Posts: 41875

Kudos [?]: 128459 [1], given: 12173

In certain city, a car is parking at A destination, it will go to B de [#permalink]

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06 Oct 2017, 00:12
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43% (01:11) correct 57% (00:31) wrong based on 21 sessions

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CHALLENGE QUESTIONS

In certain city, a car is parking at A destination, it will go to B destination along routes that are confined to square grid of x streets and y avenues, where streets and avenues are perpendicular to each other. How many routes from A to B can the car take that have the minimum possible length?

A. $$\frac{(x+y)!}{x!y!}$$

B. $$\frac{(x+y)!}{(x-1)!(y-1)!}$$

C. $$\frac{(x+y-1)!}{x!y!}$$

D. $$\frac{(x+y-2)!}{x!y!}$$

E. $$\frac{(x+y-2)!}{(x-1)!(y-1)!}$$
[Reveal] Spoiler: OA

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Kudos [?]: 128459 [1], given: 12173

Intern
Joined: 22 Dec 2016
Posts: 2

Kudos [?]: 1 [1], given: 2

Location: Kazakhstan
Concentration: Finance, General Management
WE: Analyst (Retail Banking)
Re: In certain city, a car is parking at A destination, it will go to B de [#permalink]

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06 Oct 2017, 05:06
1
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Please correct me if i'm wrong:

This is our travel map:

A.......
.........
.........
.........
.......B

In order to get from A to B we need to pass X-1 blocks (streets) plus Y-1 blocks (avenues); -1 because we start from a certain street and avenue and should not count them.
On every crossroad we can walk either DOWN or RIGHT, so our travel route will be something like DDDRDRDR, where D=X-1 times and R=Y-1 times.

We can do this in ((X-1)+(Y-1))!=(X+Y-2)! / (X-1)!(Y-1)! ways, so E?

Kudos [?]: 1 [1], given: 2

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4961

Kudos [?]: 5447 [1], given: 112

Re: In certain city, a car is parking at A destination, it will go to B de [#permalink]

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07 Oct 2017, 06:32
1
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Expert's post
Bunuel wrote:

CHALLENGE QUESTIONS

In certain city, a car is parking at A destination, it will go to B destination along routes that are confined to square grid of x streets and y avenues, where streets and avenues are perpendicular to each other. How many routes from A to B can the car take that have the minimum possible length?

A. $$\frac{(x+y)!}{x!y!}$$

B. $$\frac{(x+y)!}{(x-1)!(y-1)!}$$

C. $$\frac{(x+y-1)!}{x!y!}$$

D. $$\frac{(x+y-2)!}{x!y!}$$

E. $$\frac{(x+y-2)!}{(x-1)!(y-1)!}$$

hi..

apart from method above a QUICK solution may be..

take the simplest set of numbers - 2 streets and 2 avenues

so
12
34

ways 1-2-4 and 1-3-4, that is TWO ways

lets check choices for ans 2 when x and y are 2 each

A. $$\frac{(x+y)!}{x!y!}............\frac{(2+2)!}{2!2!}=6$$...NO

B. $$\frac{(x+y)!}{(x-1)!(y-1)!}..........\frac{4!}{1!1!}$$....NO

C. $$\frac{(x+y-1)!}{x!y!}......\frac{(2+2-1)!}{2!2!}=\frac{3}{2}.$$....NO

D. $$\frac{(x+y-2)!}{x!y!}...........\frac{(2+2-2)!}{2!2!}=\frac{1}{2}$$......NO

E. $$\frac{(x+y-2)!}{(x-1)!(y-1)!}..............\frac{(2+2-2)!}{1!1!}=2$$....YES and our answer

E
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5447 [1], given: 112

Re: In certain city, a car is parking at A destination, it will go to B de   [#permalink] 07 Oct 2017, 06:32
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