The number of multiples of $4$ between $10$ and $250$ is:

Let's find the boundaries for multiples of $4$ in this range:
* First multiple higher than $10$ is $12$.
* Last multiple lower than $250$ is $248$.

The sequential multiples form an Arithmetic Progression (AP): $12,16,20,\dots ,248$

* Initial value ($a$) = $12$
* Common difference ($d$) = $4$
* Final term ($l$) = $248$

Applying the standard formula for finding total terms ($l=a+(n-1)d$):

$$248=12+(n-1)\times 4$$

$$236=(n-1)\times 4$$

$$n-1=\frac{236}{4}=59$$

$$n=59+1=60$$