Handshake counting at a conference: if 10 people each shake hands with every other person exactly once, how many handshakes occur in total?

Introduction / Context:
This is a standard combinatorics question. Each handshake is a unique pair of people. Counting such pairs is equivalent to selecting 2 people from 10 without order, which is a combinations problem.

Given Data / Assumptions:

• Total people n = 10.
• Each unordered pair shakes hands once.
• No one shakes hands with themselves; no repeats.

Concept / Approach:
Number of unique handshakes = number of unordered pairs = nC2 = n*(n - 1)/2. Substitute n = 10 to find the count. Alternatively, use double-count logic and divide by 2 to avoid overcounting.

Step-by-Step Solution:

Verification / Alternative check:
Check by summation: first person shakes 9 hands, second shakes 8 new hands, ... down to 1; sum = 9 + 8 + … + 1 = 45. Matches the formula.

Why Other Options Are Wrong:

• 20 — too small; corresponds to 10 * 4 / 2.
• 55 — sum up to 10, not 9; would include self or double count.
• 90 — equals 10 * 9, which counts each handshake twice (ordered pairs).

Common Pitfalls:
Counting ordered pairs instead of unordered, or forgetting to divide by 2. Always ensure each handshake is counted once.

Final Answer:
45