Pascals Triangle - Number of Routes on a Grid

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Deepa's route home involves walking five blocks west and four blocks south from a bus stop, with a stop at a store located three blocks west and one block south of the bus stop. The initial calculation for the number of routes was based on the combinations of paths to the store and from the store to home, resulting in 60 ways. However, a correction was suggested, changing the combination for the route to the store to C(4,1), leading to a revised total of 40 ways. This corrected calculation was confirmed by another participant, agreeing that 40 is the accurate number of routes Deepa can take.
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Homework Statement



Deepa’s house is five blocks west and four blocks south of a bus stop. A store is three blocks west and one block south of the bus stop. How many ways can Deepa walk home if she wishes to stop on her way home by walking only west and south?

Homework Equations





The Attempt at a Solution



(number of routes to the store) * (number of routes from store to home)

= C (4,2) * C(5,2)

= (4*3/2) * (5*4/2)

= 6 * 10

= 60 ways

Therefore, Deepa can walk home in 60 ways if she wishes to stop by the store on her way home.

Would anybody be able to confirm whether my answer is correct?

Thanks
 
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crosby87 said:
(number of routes to the store) * (number of routes from store to home)
= C (4,2) * C(5,2)
Not C(4,2). Try again.
 
Must be C(4,1) then:

C (4,1) * C(5,2)

= (4) * (5*4/2)

= 4 * 10

= 40 ways

Can anyone confirm this?

Thanks!
 
40 looks right to me.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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