Let $f(x)=(c-5)x^2-2cx+(c-4), \,\, c\ne 5.$ We need integral values of $c$ such that one root lies in $(0,2)$ and the other in $(2,3)$.

1. Use sign changes across the intervals

Since the quadratic has one root in $(0,2)$ and the other in $(2,3)$, the point $x=2$ lies between the two roots.

For a quadratic, if $2$ lies between the roots, then $f(2)$ must have sign opposite to the leading coefficient $a=c-5$.

Also, to ensure one root in $(0,2)$ and the other in $(2,3)$, it is enough to require:

• $f(0)$ and $f(2)$ have opposite signs, and
• $f(2)$ and $f(3)$ have opposite signs.

Now compute:

At $x=0$

$f(0)=c-4.$

At $x=2$

$f(2)=4(c-5)-4c+(c-4)=c-24.$

At $x=3$

$f(3)=9(c-5)-6c+(c-4)=4c-49.$

So we need $(c-4)(c-24)<0$ and $(c-24)(4c-49)<0.$

2. Solve the inequalities

First inequality

$(c-4)(c-24)<0 \implies 4<c<24.$

Second inequality

$(c-24)(4c-49)<0.$ The critical points are $c=24$ and $c=\frac{49}{4}=12.25$. So, $\frac{49}{4}<c<24.$

Combining with $4<c<24$ gives $\frac{49}{4}<c<24.$

Since $c$ is integral, $c\in\{13,14,15,16,17,18,19,20,21,22,23\}.$

These are $11$ integers.

3. Check that this indeed gives one root in each interval

For every such $c$:

• $f(0)$ and $f(2)$ have opposite signs $\Rightarrow$ one root in $(0,2)$,
• $f(2)$ and $f(3)$ have opposite signs $\Rightarrow$ one root in $(2,3)$.

Since a quadratic has only two roots, these are exactly the two roots required.

Also, endpoints are excluded automatically because inequalities are strict.

4. Evaluate options

• A: $12$ — incorrect
• B: $18$ — incorrect
• C: $10$ — incorrect
• D: $11$ — correct

Final Answer

The number of elements in $S$ is $\boxed{11}.$