In a room of 6 people, how many different handshakes are possible if each person shakes hands once with every other?

Correct — C.
Rule: a handshake is an unordered pair, so count C(6, 2).

Each person shakes 5 hands: 6 × 5 = 30
That counts A–B and B–A separately, so halve it: 30 ÷ 2 = 15

Check by adding: 5 + 4 + 3 + 2 + 1 = 15 → option (c).

Why the others are wrong

• (a)
  10 — 10 is 4 + 3 + 2 + 1, the total for five people. The sixth person's 5 handshakes bring it to 15.
• (b)
  12 — 12 is 6 × 4 ÷ 2, the count if each person shook only four hands. Each shakes hands with the 5 others: 6 × 5 ÷ 2 = 15.
• (d)
  20 — 20 is C(6, 3), the number of ways to choose three people. A handshake joins two, so the count is C(6, 2) = 15.

Handshakes, round-robin matches and lines joining points are one count: pairs chosen from n, where order does not matter. The formula is n(n − 1) ÷ 2.

Build it up: each of n people meets n − 1 others, giving n(n − 1), but every handshake is counted once from each end, so divide by 2.

For 6 people: 6 × 5 ÷ 2 = 15.

Key facts

• Handshakes among n people = n(n − 1) ÷ 2.
• For n = 6: 6 × 5 ÷ 2 = 15.
• The same count is 1 + 2 + … + (n − 1): for 6 people, 1 + 2 + 3 + 4 + 5 = 15.

Common traps

• Forgetting to halve n(n − 1), which counts every handshake twice.
• Using C(6, 3) = 20, which counts groups of three, not pairs.

No directly related past PYQ was found.