Counting directions

In order to arrive at point $P$ after $6$ steps, the net movement in any direction North, South and East, West must be zero. We can derive the following system of equations:

Let $N,S,E,W$ represent the number of steps taken in directions North, South, East and West respectively:

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$\begin{aligned} N-S &= 0 \quad \text{Eq1} \\ \quad \quad \\ E-W &= 0 \quad \quad \text{Eq2} \\ \quad \quad \\ N + S + E + W &= 6 \quad \quad \text{Eq3} \\ \end{aligned}$

$\Rightarrow$

$\begin{aligned} N &= S \quad \quad \text{Eq1'} \\ \quad \\ E &= W \quad \quad \text{Eq2'} \\ \quad \\ & \quad \quad \text{Sub Eq1' \&} \, \text{Eq2'} \to \text{Eq3} \\ \quad \\ N + E &= 3 \quad \quad \text{Eq3'} \\ \end{aligned}$

We need to find solutions to $\text{Eq3'} \in \mathbf{N}$

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There are $4$ solutions:

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$\begin{array}{c}&N \quad &S \quad &E \quad &W \\ &0 \quad &0 \quad &3 \quad &3 \\ &3 \quad &3 \quad &0 \quad &0 \\ &1 \quad &1 \quad &2 \quad &2 \\ &2 \quad &2 \quad &1 \quad &1 \end{array}$

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The number of paths $N_s( N,S,E,W )$ are counted by arranging the number of steps for each a solution as a string of $6$ letters. Examples of such a sequence of steps might be:

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$N,E,W,W,E,S$ or $S,S,E,N,N,W$ etc...

$\quad$

The results of the possible orderings for the solution set are as follows:

$\begin{aligned} N_s ( 0,0,3,3) & = \overbrace{ { 6 \choose 3} }^{ \scriptsize \begin{array}{c}\text{3 E's} \\ \text{in} \\ \text{6 places} \end{array}} \cdot \overbrace{ { 3 \choose 3} }^{ \scriptsize \begin{array}{c}\text{3 W's} \\ \text{in} \\ \text{3 places} \end{array}} = 20 \\ \quad \\ N_s ( 3,3,0,0) &= \overbrace{ { 6 \choose 3} }^{ \scriptsize \begin{array}{c}\text{3 N's} \\ \text{in} \\ \text{6 places} \end{array}} \cdot \overbrace{ { 3 \choose 3} }^{ \scriptsize \begin{array}{c}\text{3 S's} \\ \text{in} \\ \text{3 places} \end{array}} = 20 \\ \quad \\ N_s ( 1,1,2,2) & = \overbrace{ { 6 \choose 1} }^{ \scriptsize \begin{array}{c}\text{1 N} \\ \text{in} \\ \text{6 places} \end{array} } \cdot \overbrace{ { 5 \choose 1} }^{\scriptsize \begin{array}{c}\text{ 1 S} \\ \text{in} \\ \text{5 places} \end{array} } \cdot \overbrace{ { 4 \choose 2} }^{ \scriptsize \begin{array}{c}\text{2E's} \\ \text{in} \\ \text{4 places} \end{array}} \cdot \overbrace{ { 2 \choose 2} }^{\scriptsize \begin{array}{c}\text{2 W's} \\ \text{in} \\ \text{2 places} \end{array}} = 180 \\ \quad \\ N_s ( 2,2,1,1) & = \overbrace{ { 6 \choose 2} }^{ \scriptsize \begin{array}{c}\text{2 N's} \\ \text{in} \\ \text{6 places} \end{array} } \cdot \overbrace{ { 4 \choose 2} }^{\scriptsize \begin{array}{c}\text{2 S's} \\ \text{in} \\ \text{4 places} \end{array} } \cdot \overbrace{ { 2 \choose 1} }^{ \scriptsize \begin{array}{c}\text{1 E} \\ \text{in} \\ \text{2 places} \end{array}} \cdot \overbrace{ { 1 \choose 1} }^{\scriptsize \begin{array}{c}\text{1 W} \\ \text{in} \\ \text{1 place} \end{array}} = 180 \end{aligned}$

The number of paths from $P$ back to $P$ in exactly $6$ steps is then:

$2 \cdot ( 20 + 180 ) = 400$