The problem involves determining how many people are present in the room if the total number of handshakes is 66.

The number of handshakes between $n$ people can be determined by the formula for combinations:

$\text{Total Handshakes} = \binom{n}{2} = \frac{n(n-1)}{2}$

We are given that the total number of handshakes is 66, so:

$\frac{n(n-1)}{2} = 66$

To solve for $n$, first multiply both sides by 2 to eliminate the denominator:

$n(n-1) = 132$

Now solve the quadratic equation:

$n^2 - n - 132 = 0$

Using the quadratic formula, $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -1$, and $c = -132$:

$n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-132)}}{2(1)}$

$n = \frac{1 \pm \sqrt{1 + 528}}{2}$

$n = \frac{1 \pm \sqrt{529}}{2}$

$n = \frac{1 \pm 23}{2}$

Thus, $n = \frac{1 + 23}{2} = 12$ or $n = \frac{1 - 23}{2} = -11$.

Since $n$ must be positive, the total number of people is $n = 12$.

Answer:

There are 12 people present in the room.

Would you like further details or have any questions about this solution?

Here are 5 related questions for further exploration:

1. What would the number of handshakes be if there were 15 people?
2. How would the formula change if two people did not shake hands with each other?
3. Can you find a general formula for the total number of handshakes for $n$ people?
4. If one person left the room, how many handshakes would occur?
5. How can this handshake problem be represented as a graph?

Tip: For any group of $n$ people, the number of interactions or combinations can be calculated using $\binom{n}{2}$, which simplifies counting pairwise interactions like handshakes.