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Bunuel
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In certain city, a car is parking at A destination, it will go to B de 06 Oct 2017, 00:12
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Difficulty:

95% (hard)

Question Stats:

32% (02:02) correct 68% (01:33) wrong based on 111 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. \(\frac{(x+y-2)!}{(x-1)!(y-1)!}\)
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E
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In certain city, a car is parking at A destination, it will go to B de 06 Oct 2017, 05:06
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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?
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In certain city, a car is parking at A destination, it will go to B de 07 Oct 2017, 06:32
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Bunuel

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. \(\frac{(x+y-2)!}{(x-1)!(y-1)!}\)
Show more


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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In certain city, a car is parking at A destination, it will go to B de 27 Oct 2017, 02:06
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chetan2u
Bunuel

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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Hi chetan2u
can you draw a picture to explain your way, please? unfortunately, I can't imagine this question
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In certain city, a car is parking at A destination, it will go to B de 27 Oct 2017, 05:59
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soodia
chetan2u
Bunuel

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



Hi chetan2u
can you draw a picture to explain your way, please? unfortunately, I can't imagine this question
Show more

Hi,

here is a sketch that may help you
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streets.png
streets.png [ 10.69 KiB | Viewed 2936 times ]

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In certain city, a car is parking at A destination, it will go to B de 29 Oct 2023, 02:14
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