Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be the function given by
$f(x) = |x - 2| + 3|x - 1| + ||x - 2| - 1|$.
What is the number of points where $f$ is NOT differentiable?

The given function is $f(x) = |x - 2| + 3|x - 1| + ||x - 2| - 1|$.

The possible points of non-differentiability are the critical points where the expressions inside the absolute values become zero:
$x - 2 = 0 \Rightarrow x = 2$
$x - 1 = 0 \Rightarrow x = 1$
$|x - 2| - 1 = 0 \Rightarrow |x - 2| = 1 \Rightarrow x = 1$ or $x = 3$

Thus, the critical points to check are $x = 1, 2, 3$. Let us define $f(x)$ in the intervals around these points.

For $x < 1$:
$f(x) = (2 - x) - 3(x - 1) + (2 - x - 1)$
$f(x) = 2 - x - 3x + 3 + 1 - x = 6 - 5x$
The derivative is $f'(x) = -5$.

For $1 < x < 2$:
$f(x) = (2 - x) + 3(x - 1) - (2 - x - 1)$
$f(x) = 2 - x + 3x - 3 - 1 + x = 3x - 2$
The derivative is $f'(x) = 3$.

For $2 < x < 3$:
$f(x) = (x - 2) + 3(x - 1) - (x - 2 - 1)$
$f(x) = x - 2 + 3x - 3 - x + 3 = 3x - 2$
The derivative is $f'(x) = 3$.

For $x > 3$:
$f(x) = (x - 2) + 3(x - 1) + (x - 2 - 1)$
$f(x) = x - 2 + 3x - 3 + x - 3 = 5x - 8$
The derivative is $f'(x) = 5$.

Now, checking the differentiability at the critical points:
At $x = 1$: Left Hand Derivative (LHD) $= -5$, Right Hand Derivative (RHD) $= 3$. Since LHD $\neq$ RHD, $f(x)$ is not differentiable at $x = 1$.
At $x = 2$: LHD $= 3$, RHD $= 3$. Since LHD $=$ RHD, $f(x)$ is differentiable at $x = 2$.
At $x = 3$: LHD $= 3$, RHD $= 5$. Since LHD $\neq$ RHD, $f(x)$ is not differentiable at $x = 3$.

Therefore, there are exactly $2$ points ($x = 1$ and $x = 3$) where the function is not differentiable.

Answer: $2$