Chris and his wife invited a total of ten families on their marriage anniversary. While the host family had just the two members, each family invited consisted of four members. If every person in the party shook hands exactly once with every other person belonging to a different family, then find the total number of handshakes that took place in the party.

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729

800

Correct

801

850

851

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Explanation

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Official Explanation

Official

We have ten families of 4, and one family of 2, so there are 42 people in total.

If everyone shook hands with everyone else, then every pair of people would produce a handshake, and we have

\binom{42}{2} = \frac{42 \times 41}{2!} = 861

pairs of people.

But we know that within each family, no one shakes hands. In the two-person family, that means 1 handshake doesn't take place, and within each of the ten four-person families,

\binom{4}{2} = 6

handshakes don't take place.

So there are

1 + 10 \times 6 = 61

handshakes we shouldn't count, and the answer is

861 - 61 = 800

.…

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What this question tests

What it tests

Whether you can count arrangements and selections correctly — permutations vs. combinations.

Common trap

Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.