Consider the quadratic equation (c - 5) x ^ 2 - 2cx + (c - 4) = 0,...

Case $- I$

$c - 5 > 0$

$⇒ c > 5 . . . \left(i\right)$

Then, the graph of the quadratic is shown below

Hence, we have $f\left(0\right) > 0$

$⇒ c - 4 > 0$

$⇒ c > 4 . . . \left(ii\right)$

And $f\left(2\right) < 0$

$⇒ 4\left(c - 5\right) - 4c + c - 4 < 0$

$⇒ 4c - 20 - 4c + c - 4 < 0$

$⇒ c < 24 . . . \left(iii\right)$

And $f\left(3\right) > 0$

$⇒ 9\left(c - 5\right) - 6c + c - 4 > 0$

$⇒ 9c - 45 - 6c + c - 4 > 0$

$⇒ 4c - 49 > 0$

$⇒ c > \dfrac{49}{4} . . . \left(iv\right)$

Hence, on taking the intersection of all the above inequalities, we get $c \in \left(\dfrac{49}{4}, 24\right) .$

Case $- II$

$c - 5 < 0$

$⇒ c < 5 . . . \left(v\right)$

Then, the graph of the quadratic is shown below

Hence, we have $f\left(0\right) < 0$

$⇒ c < 4 . . . \left(vi\right)$

And, $f\left(2\right) > 0$

$⇒ c > 24 . . . \left(vii\right)$

Also, $f\left(3\right) < 0$

$⇒ c < \dfrac{49}{4} . . . \left(viii\right)$

Hence, on taking the intersection of all the above inequalities, we get $⇒ c \in ϕ$

Hence, $c \in \left(\dfrac{49}{4}, 24\right)$

Hence, the total number of element in $S$ are $11 .$