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**+1 **

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

CPhill
Jul 10, 2018