Initially, there are $$3^{80}$$ particles at the origin $$(0, 0)$$. At each step the particles are moved to points above the x-axis as follows: if there are $$n$$ particles at any point $$(x, y)$$, then $$\frac{n}{3}$$ of them are moved to $$(x+1,y+1)$$, $$\frac{n}{3}$$ are moved to $$(x,y+1)$$ and the remaining to $$(x-1,y+1)$$. For example, after the first step, there are $$3^{79}$$ particles each at $$(1, 1)$$, $$(0, 1)$$ and $$(-1, 1)$$. After the second step, there are $$3^{78}$$ particles each at $$(2, 2)$$ and $$(-2, 2)$$, $$2\cdot 3^{78}$$ particles each at $$(1, 2)$$ and $$(-1, 2)$$, and $$3^{79}$$ particles at $$(0, 2)$$. After $$80$$ steps, the number of particles at $$(79, 80)$$ is:

Each particle performs exactly $$80$$ moves. At every move it can go to
$$x+1,\;x,\;x-1$$ while $$y$$ definitely increases by $$1$$. Thus after $$80$$ moves every particle is somewhere on the line $$y=80$$ and its final $$x$$-coordinate is the algebraic sum of the $$80$$ chosen steps $$\left(+1,0,-1\right)$$.

The splitting rule ensures that every particle chooses one of the three options with equal proportion. Because the origin initially holds $$3^{80}$$ particles and the branching factor is $$3$$ at every step, the total number of distinct step-sequences of length $$80$$ is also $$3^{80}$$. Hence exactly one particle follows each possible sequence.
Consequently, the number of particles that finally reach any particular point $$(x,80)$$ equals the number of step-sequences that sum to that $$x$$.

Let
$$a=$$ number of $$+1$$ steps,  $$b=$$ number of $$0$$ steps,  $$c=$$ number of $$-1$$ steps.

We require two equations:
1. Total steps: $$a+b+c=80$$  $$-(1)$$
2. Final $$x$$-coordinate: $$a-c=79$$  $$-(2)$$

From $$-(2)$$, $$c=a-79\,. $$ Substitute this in $$-(1)$$:

$$a+b+(a-79)=80 \;\Longrightarrow\; b=159-2a.$$

All variables must be non-negative integers.

Because $$c=a-79\ge0$$, we need $$a\ge79$$. Trying $$a=79$$ gives $$c=0,\; b=159-2\!\times\!79=1,$$ which is admissible.
Trying $$a=80$$ gives $$c=1,\; b=159-160=-1,$$ which is impossible.

Therefore the only feasible distribution is
$$a=79,\; b=1,\; c=0.$$/p>

This means every valid sequence consists of exactly $$79$$ moves of $$+1$$ and exactly $$1$$ move of $$0$$, in some order. The position of the lone $$0$$ can be chosen in $$\binom{80}{1}=80$$ ways.

Hence, there are $$80$$ distinct step-sequences that end at $$(79,80)$$, and so the number of particles present there after $$80$$ steps is also $$80$$.

Final Answer: 80