Probability of choosing envelopes 
Suppose that you have 20 different letters and 10 distinctly addressed envelopes. The 20 letters consists of 10 pairs, where each pair belongs inside one of the 10 envelopes. Suppose that you place the 20 letters inside the 10 envelopes, two per envelope, but at random.
What is the probability that exactly 3 of the 10 envelopes will contain both of the letters which they should contain?

I have seen similar questions to this one but they always assign only one letter to one envelope. Also, the scenario is usually how to choose AT LEAST one right envelope. I am not too clear as to how to adapt to this new scenario.

Let $S_i$ be the arrangements where envelope $i$ has both of its intended letters. The number of intersections of $k$ of the $S_i$ is
$$
N_k=\overbrace{\ \ \binom{10}{k}\ \ }^{\substack{\text{number of ways}\\\text{to choose the}\\\text{$k$ envelopes}}}\ \ \overbrace{\frac{(20-2k)!}{2^{10-k}}\vphantom{\binom{10}{k}}}^{\substack{\text{ways to arrange}\\\text{$20-2k$ letters in}\\\text{$10-k$ envelopes}}}
$$
The Generalized Principle of Inclusion-Exclusion says that the number of arrangements with exactly $3$ envelopes properly filled is
$$
\begin{align}
\sum_{k=3}^{10}(-1)^{k-3}\binom{k}{3}\binom{10}{k}\frac{(20-2k)!}{2^{10-k}}
&=\binom{10}{3}\sum_{k=3}^{10}(-1)^{k-3}\binom{7}{k-3}\frac{(20-2k)!}{2^{10-k}}\\[6pt]
&={75718299600}
\end{align}
$$
The number of ways to arrange $20$ letters in $10$ envelopes is
$$
\frac{20!}{2^{10}}={2375880867360000}
$$
Thus, the requested probability is
$$
\begin{align}
\frac{75718299600}{2375880867360000}
&=\frac{21032861}{659966907600}\\[6pt]
&\approx0.0000318695691523
\end{align}
$$
