All-Soviet-Union Mathematical Olympiad

Solution:

a. Each meeting involves $\frac{10 \cdot 9}{2} = 45$ pairs. So after 40 meetings, there have been $40 \times 45 = 1800$ pairs. We are told that these are all distinct. But if there are $N$ people on the committee, then there are only $\frac{N(N-1)}{2}$ pairs available. For $N=60$, this is only $\frac{60 \cdot 59}{2} = 1770$. Therefore, $N > 60$.

b. A subcommittee of 5 has $\frac{5 \cdot 4}{2} = 10$ pairs. So 31 subcommittees have $31 \times 10 = 310$ pairs, and these are all distinct, since no two people are on more than one subcommittee. But a committee of 25 only has $\frac{25 \cdot 24}{2} = 300$ pairs available. Therefore, you cannot make more than 30 such subcommittees.