Each of the 31 people shook hands with 30 other people. So it should be 930 handshakes, right? No, because each handshake involves 2 people, thus we have double-counted:

$\dfrac{930}{2} = \boxed{465}$

Solution 1 - Combinatorics

In this group of people, each pair of people will handshake exactly once. Thus, we just need to count how many pairs of people there are in this group, which can be easily done by calculating $31$ choose $2$ . $_{31}C_{2}=\boxed{465}$ .

Solution 2 - Overcounting

Let's pretend that you are one of the $31$ people in the room. Because you will have to handshake the remaining people in the room, you will perform exactly $30$ handshakes. This will apply to every student in the room, so in total, every student will count $31 \times 30$ handshakes. However, each handshake was counted twice: once by the person on the left and once by the person on the right. Thus, we have to divide by $2$ . This gives us $\frac{31 \times 30}{2}=\boxed{465}$ .

Practical Solution (It's a joke.)

There will be $0$ handshakes due to COVID - $19$ .

n = 2, there is 1 handshake

next guest comes,

n =3, s/he need to make a handshake to each previous two guests, so total = 1 + 2

n =4, s/he need to make a handshake to each previous three guests, so total = 1 + 2 + 3

so when n = 31, there will have 1 + 2 + 3 +.... + 30 handshakes = $\boxed{465}$

There is a formula for $n$ handshakes which is : $\frac{n(n-1)}{2}=\frac{31\times30}{2}=465$ ; in our case $n=31$

First person $\rightarrow 30$ Second person $\rightarrow 29$ Third person $\rightarrow 28 \cdots$

$\therefore$ Total $=30+29+28 \cdots +1 = \dfrac{30(31)}{2}=\boxed{465}$