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 avenuesso

12

34

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

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