probability of exactly 3 people choosing the same two numbers from n numbers 
There are 30 people. Each of them will choose randomly two distinct numbers from the set $\{1,2,\cdots,20\}$.
What is the probability that exactly 3 of them will choose the same two numbers?

Let A denote the event that exactly 3 of them will choose the same two numbers.
Note that we have $\binom{20}{2}=190$ different ways to choose 2 distinct numbers from $\{1,2,\cdots,20\}$.
I see that there are basically two cases: all the remaining 27 choose distinct numbers or there might be groups of 2 people with the same numbers. I have difficulty with the second case!
Therefore, $P(A)=\frac{\binom{190}{1} \binom{30}{3}\binom{189}{27}27!+(???)}{190^{30}}$

If you have $189$ selections that can be made by $27$ people with $p$ selections each chosen by two people  and $27-2p$ each chosen by one person, there are $\frac{189!\,27!}{p!\,(27-2p)!\,(162+p)!\,2^p}$ ways of doing that.
Sum that over $0\le p \le 13$ and you get about $2.7025\times 10^{61}$
Multiply that by your  $\binom{190}{1} \binom{30}{3}$ and you get about $2.0847\times 10^{67}$
Divide that by $190^{30}$ and you get a probability of about $0.09046$.  Simulation suggests this is sensible.
Doing something similar with $190$ selections and $30$ people with no triples and you would get a probability of a less extreme result of about $0.90257$, leaving about $0.00697$ for a more extreme result
