The problem involves finding the number of handshakes between people where each person shakes hands with every other person exactly once.

To solve this, we use the concept of combinations. If there are $n$ people, each person can shake hands with $n-1$ others, but we must avoid counting the same handshake twice. Therefore, the number of handshakes is given by the combination formula:

$\text{Number of handshakes} = \binom{n}{2} = \frac{n(n-1)}{2}$

In this case, there are 4 people:

$\text{Number of handshakes} = \binom{4}{2} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$

So, there are 6 handshakes.

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How many handshakes would there be if there were 5 people in the room?
2. What is the number of handshakes if there are 10 people?
3. How does the formula for combinations apply to this handshake problem?
4. What if each person in the room shakes hands twice with every other person? How does that change the result?
5. How would you approach this problem if there were 100 people?

Tip: The combination formula $\binom{n}{r}$ is useful in counting problems where order does not matter, like this handshake scenario.