Right up and away

First, with $r_{k}$ representing the length of the $k$ th right step and $u_{j}$ representing the length of the $j$ th up step, we need to look at the solutions of the equation

$r_{1} + u_{1} + r_{2} + u_{2} + r_{3} + u_{3} = 10.$

Now since each of these variables must be $\ge 1,$ we can let each $r_{k}' = r_{k} - 1$ and $u_{j}' = u_{j} - 1,$ and instead look at the equation

$r_{1}' + u_{1}' + r_{2}' + u_{2}' + r_{3}' + u_{3}' = 4,$

where each of the variables is non-negative. This is a stars and bars equation with solution $\dbinom{9}{4} = 126.$

Now what matters here as far as calculating $D$ is concerned is the total distances right and up the particle moves after six steps. So we can have $\sum_{k=1}^{3} r_{k}' = 0,1,2,3,4$ with corresponding values $\sum_{j=0}^{4} u_{j}' = 4,3,2,1,0.$ So now we have a series of additional stars and bars equations to solve. For example, with the sum of the $r_{k}'$ s being $3$ , we are solving the equation $r_{1}' + r_{2}' + r_{3}' = 3,$ which has the solution $\binom{5}{3} = 10.$ The corresponding $u_{j}$ equation is $u_{1}' + u_{2}' + u_{3}' = 1,$ which has solution $\binom{3}{1} = 3.$ Thus there are $10*3 = 30$ paths that, after the six steps are completed, involve the particle going $6$ units right and $4$ units up, which corresponds to a distance of $\sqrt{6^{2} + 4^{2}} = 2\sqrt{13}.$

Similarly, there are:

• $15$ paths each where the particle either moves $7$ units to the right and $3$ units up, or $3$ units to the right and $7$ units up, resulting in a distance of $\sqrt{49 + 9} = \sqrt{58}$ ;

• $30$ paths each where the particle either moves $6$ units right and $4$ units up, or $4$ units right and $6$ units up, resulting in a distance of $2\sqrt{13}$ ;

• $36$ paths where the particle goes both $5$ units right and $5$ units up, for a distance of $5\sqrt{2}.$

Thus $D = \dfrac{30}{126}\sqrt{58} + \dfrac{60}{126}(2\sqrt{13}) + \dfrac{36}{126}(5\sqrt{2}) =$

$\dfrac{5}{21}\sqrt{58} + \dfrac{20}{21}\sqrt{13} + \dfrac{10}{7}\sqrt{2} = 7.26744....$

Thus $\lfloor 1000*D \rfloor = \boxed{7267}.$

Thanks to Brian for helping me fix this.

By stars and bars, we have the number of ways to go 3-7 units right and 7-3 units up is

$\displaystyle \sum_2^6 \dbinom{n}{2}\dbinom{8-n}{2}=126$

The total length traveled by each of these will be

$\sqrt{(n+1)^2+(9-n)^2}$

So we evaluate the sum and divide by the total number of cases to get to be (note that we begin the sum at n=2 and not 3 so that's why the sqrt changes)

$\dfrac{\displaystyle \sum_2^6 \dbinom{n}{2}\dbinom{8-n}{2}\sqrt{(n+1)^2+(9-n)^2}}{\displaystyle \sum_2^6 \dbinom{n}{2}\dbinom{8-n}{2}}\approx 7.397$