Formalities

Let the number of men be $m$ and the number of women be $w$ . We will now construct a statement to represent the number of handshakes that occurred.

For a man, he must shake hands with every other man twice. There are $m$ men that must do this, but we double count since a handshake between $m_a$ and $m_b$ is the same as a handshake between $m_b$ and $m_a$ . Thus, the expression for this is

$\dfrac {m(m-1)}{2} \times 2 = m(m-1)$

Also, each man must shake hands with each woman once. This expression is $mw$ . Adding these two expressions, we get a total of $m(m-1)+mw = m(m+w-1)$ handshakes. We already know $m+w=37$ and $m(m+w-1)=720$ . Thus, $m+w-1=36$ so $m=20$ , from which we get $w=17$ .

Let the number of men in this event be $m$ . How each man shake hands each other once when the event start, we have to find the combinations:

$\left( \begin{matrix} m \\ 2 \end{matrix} \right) =\frac { m! }{ 2!(m-2)! }$

How it repeats in the end of the event, we need to double it:

$2\cdot \frac { m! }{ 2!(m-2)! }$

The men shake hands with each women when the event starts. How a man shake hands with a woman once, we multiply the number of men $m$ and the number of women $37-m$ resulting in $37m-m^2$ . Add up all of this we will get the total of "shake hands" which is 720:

$2\cdot \frac { m! }{ 2!(m-2)! } +37m-{ m }^{ 2 }=720$

Simplifying:

$\frac { m! }{ (m-2)! } +37m-{ m }^{ 2 }=720$

$\frac { m(m-1)((m-2)!) }{ (m-2)! } +37m-{ m }^{ 2 }=720$

${ m }^{ 2 }-m+37m-{ m }^{ 2 }=720$

$36m=720\quad \therefore \quad m=20$

Then, there were 20 men in this event. It mean that there were $37-20=17$ women.