Number of integral values of $c$ in solution set 
Let the quadratic equation $(c-5)x^2-2cx+c-4=0$
has one root in $(0,2)$ and other root in $(2,3).$
Then the number of integral values of $c$ in solution set

Try: writing quadratic equation as $$f(x)=x^2-\bigg(\frac{2c}{c-5}\bigg)x+\frac{c-4}{c-5}=0\;\;, c\neq 5$$
$f(x)$ is upward parabola which cut $x$ axis at $(0,2)$ and other intersection in $(2,3)$
i. e $x=2$ lie between the roots means $f(2)<0$
$$f(2)=4-\frac{4c}{c-5}+\frac{c-4}{c-5}<0$$
$$\frac{4(c-5)-4c+c-4}{c-5}<0$$
$$\frac{c-24}{c-5}<0\Rightarrow 5<c<24$$
I am getting integer values of $c$ are $18$
but answer given as $11$
could someone  help me whats wrong in my reasoning

You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
