Handshake Calculation for 11 Persons

This problem involves finding the total number of unique handshakes within a group of 11 individuals, where each person shakes hands with every other person exactly once. This is a classic combinatorics problem.

Identifying the Problem Type

A handshake involves two people. Since shaking hands is mutual (Person A shaking hands with Person B is the same handshake as Person B with Person A), the order does not matter. Therefore, we need to use combinations to find the number of ways to choose 2 people from the group of 11.

Combination Formula

The number of combinations of choosing $k$ items from a set of $n$ items is given by the formula:

$C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}$

For the handshake problem, we are choosing pairs of people ($k=2$) from the total group ($n=11$).

Alternatively, a simplified formula for the specific case of handshakes ($k=2$) is:

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

Calculation Steps

1. Identify the total number of persons: $n = 11$.
2. Apply the handshake formula: Substitute $n=11$ into the formula $\frac{n(n-1)}{2}$.
3. Calculate the result: $ \text{Total Handshakes} = \frac{11 \times (11-1)}{2} $ $ \text{Total Handshakes} = \frac{11 \times 10}{2} $ $ \text{Total Handshakes} = \frac{110}{2} $ $ \text{Total Handshakes} = 55 $

Conclusion

The total number of unique handshakes possible in a group of 11 persons, where each person shakes hands with every other person exactly once, is 55.