- No set of 3 mutual acquaintances
- A person has odd number of acquaintances
- Someone knows all 26 others
- Someone knows exactly 10 others
- Everyone has a different number of acquaintances
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Answer: E
Approach: Pigeonhole on the number of acquaintances (0 to 26) among 27 people.
Solution:
There are 13 + 14 = 27 people; each person can know 0 to 26 others — 27 possible values.
If all 27 people had different counts, the counts would be exactly 0, 1, 2, …, 26.
Then someone knows all 26 others, and someone knows nobody — but the person who knows everyone knows that person too. Contradiction.
So 'everyone has a different number of acquaintances' can never be true.
Why the other options are wrong:
A: possible, e.g. nobody knows anyone.
B: possible, e.g. two people know only each other (1 each).
C: possible, e.g. one host knows all 26 guests.
D: possible, e.g. one person knows exactly 10 others.
Cetking shortcut: 0 and n − 1 can never coexist, so n people have at most n − 1 distinct counts.
Answer: E