Question: Concerning Simplifying Random Walk from 2D to 1D

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I have a question that has been confusing me. For a 1D random walk in the x-direction I was told that the multinomial coefficient is given by:

$$C(N,k_x) = \frac{N!}{k_x!(N-k_x)!} \tag{1}$$

In Eq. 1, N is the total number of steps taken, and $k_x$ is the number of steps taken to the right. And that for a 2D random walk in the x-y plane I was told that the multinomial coefficient is given by:

$$C(N,k_x, k_y) = \frac{N!}{k_x!k_y!} \tag{2}$$

So what confuses me is that it doesn't seem like the 2D random walk simplifies to the same equation used by the 1D random walk when $k_y=0$ (that is when its forced to be confined to a line in the plane). I was expecting that to have to be the case. For example, if $k_y=0$ then:

$$C(N,k_x, k_y=0) = \frac{N!}{k_x!0!}=\frac{N!}{k_x!}\neq \frac{N!}{k_x!(N-k_x)!} \tag{3}$$

I must be missing something obvious as this can not possibly be the case. What am I not understanding here?