A man throws a fair coin repeatedly. He gets $10$ points for each head he throws and $5$ points for each tail he throws. If the probability that he gets exactly $30$ points is $\dfrac{m}{n}$, $\gcd(m, n) = 1$, then $m + n$ is equal to:

Let $h$ be the number of heads and $t$ be the number of tails obtained.

The total points obtained is $10h + 5t$.

For the man to get exactly $30$ points, we must have:

$10h + 5t = 30 \Rightarrow 2h + t = 6$

Since $h$ and $t$ are non-negative integers, the possible pairs of $(h, t)$ are $(3, 0)$, $(2, 2)$, $(1, 4)$, and $(0, 6)$.

For each pair, the number of possible sequences of heads and tails is given by $^{h+t}C_{h}$, and the probability of any specific sequence is $\left(\dfrac{1}{2}\right)^{h+t}$.

The total probability is the sum of the probabilities of all these mutually exclusive cases:

For $h = 3, t = 0$, the probability is $^{3}C_{3} \left(\dfrac{1}{2}\right)^{3} = 1 \times \dfrac{1}{8} = \dfrac{8}{64}$.

For $h = 2, t = 2$, the probability is $^{4}C_{2} \left(\dfrac{1}{2}\right)^{4} = 6 \times \dfrac{1}{16} = \dfrac{24}{64}$.

For $h = 1, t = 4$, the probability is $^{5}C_{1} \left(\dfrac{1}{2}\right)^{5} = 5 \times \dfrac{1}{32} = \dfrac{10}{64}$.

For $h = 0, t = 6$, the probability is $^{6}C_{0} \left(\dfrac{1}{2}\right)^{6} = 1 \times \dfrac{1}{64} = \dfrac{1}{64}$.

Total probability $= \dfrac{8}{64} + \dfrac{24}{64} + \dfrac{10}{64} + \dfrac{1}{64} = \dfrac{43}{64}$.

Thus, $m = 43$ and $n = 64$.

Since $\gcd(43, 64) = 1$, the fraction is in its simplest form.

$m + n = 43 + 64 = 107$.

Answer: $107$