Jeff has a deck of $12$ cards: $4$ $L$s, $4$ $M$s, and $4$ $T$s. Armaan randomly draws three cards without replacement. The probability that he takes $3$ $L$s can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m +n$.
1. First, we need to determine the probability of drawing an $L$ on the first draw. Since there are 4 $L$ cards out of a total of 12 cards, the probability is:
   \[
   P(\text{First } L) = \frac{4}{12} = \frac{1}{3}
   \]

2. Next, we need to determine the probability of drawing an $L$ on the second draw, given that the first card drawn was an $L$. Now, there are 3 $L$ cards left out of a total of 11 cards:
   \[
   P(\text{Second } L \mid \text{First } L) = \frac{3}{11}
   \]

3. Finally, we need to determine the probability of drawing an $L$ on the third draw, given that the first two cards drawn were both $L$. Now, there are 2 $L$ cards left out of a total of 10 cards:
   \[
   P(\text{Third } L \mid \text{First and Second } L) = \frac{2}{10} = \frac{1}{5}
   \]

4. The probability of drawing 3 $L$ cards in succession is the product of these individual probabilities:
   \[
   P(\text{3 } L \text{s}) = \frac{1}{3} \cdot \frac{3}{11} \cdot \frac{1}{5} = \frac{1 \cdot 3 \cdot 1}{3 \cdot 11 \cdot 5} = \frac{3}{165} = \frac{1}{55}
   \]

5. The probability that Armaan draws 3 $L$ cards can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Here, $m = 1$ and $n = 55$.

6. Therefore, $m + n = 1 + 55 = 56$.

The final answer is $\boxed{56}$