At a party there are 12 people and each person shakes hands exactly once with every other person present. How many distinct handshakes take place in total?

Introduction / Context:
This is a classic handshake problem that uses combinations. It checks whether you can recognize that each handshake involves a unique pair of people and then count how many such pairs exist in a group of 12 people. Problems like this often appear in aptitude tests to assess understanding of basic combinatorics rather than brute force counting.

Given Data / Assumptions:
Number of people at the party = 12. Each handshake occurs between exactly two different people. Any pair of people shakes hands at most once. No person shakes hands with themselves.

Concept / Approach:
A handshake is uniquely determined by choosing a pair of distinct people. Therefore, the total number of handshakes is the number of ways of selecting 2 people out of 12. This is a combination because the order of people in a handshake does not matter. The required count is given by the binomial coefficient nC2, where n is the number of people. Here n = 12, so we calculate 12C2.

Step-by-Step Solution:
Step 1: Recognize that each handshake corresponds to a pair of people. Step 2: Use the combination formula nC2 = n * (n - 1) / 2. Step 3: Substitute n = 12 to get 12C2 = 12 * 11 / 2. Step 4: Compute 12 * 11 = 132. Step 5: Divide by 2 to get 132 / 2 = 66. Step 6: Therefore, the total number of distinct handshakes is 66.

Verification / Alternative check:
Another way to think of the situation is to imagine each of the 12 people counting how many other people they can shake hands with. Each person can shake hands with 11 others, so if you multiply 12 * 11 you get 132. However, this counts every handshake twice (once from each person's perspective). To correct for this double counting, divide by 2, which again gives 132 / 2 = 66 handshakes. This matches the combination approach and confirms that the answer is consistent.

Why Other Options Are Wrong:
Values like 72, 76, 64 and 78 arise from incorrectly multiplying without dividing by 2, or from random guesses. For example, 12 * 6 = 72 is a common incorrect idea where someone assumes each person shakes hands with 6 people, which is not stated in the problem. None of these alternatives are equal to 12C2. Only 66 correctly reflects the number of unique pairs in a group of 12 people.

Common Pitfalls:
Students often forget to divide by 2 when using the approach of total handshakes counted from each person, leading to 132 instead of 66. Others might mistakenly treat the problem as permutations, incorrectly using nP2 rather than nC2. Some might even attempt to list handshakes manually and lose track or double count. Remember that combinations are used when order does not matter, which is exactly the case for handshakes.

Final Answer:
The total number of distinct handshakes that take place at the party is 66.