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How many ways are there to walk from \(A\) to \(B\) by only moving rightward or upward?

Here's what I did to get my (incorrect) answer.

To get from \(A\) to \(B\), you have to move six up and six right.  \(6+6=12 \\ \binom{12}{6}=\color{red}{\boxed{924}\leftarrow\text{INCORRECT}}\)

Guest Jul 10, 2018
 #1
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Note that this just boils down to counting the arrangements  of  sets

 

To get from A to the intermediate point  we can go

(East, East, North, North, North)

 

And  the total number of arrangements of this set  is  just place East  in any of the two of the  five positions  (or, alternatively, North in any of three of the five positions)...so  C(5,2)   = C (5,3)   = 10  ways

 

Next...to get from the intermediate point to B we have

 

(East, East, East , East, North , North , North)

So....the possible arangements of this set  is either placing East  in any four of the seven positions or to place North in any of three of the seven positions  =   C( 7,4)  =  C(7,3)   =   35  ways

 

So...the number of ways to  get from A to B  =   10 * 35   = 350  ways

 

 

cool cool cool

CPhill  Jul 10, 2018

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