# 1959 AHSME Problems/Problem 50

A club with $x$ members is organized into four committees in accordance with these two rules: $$\text{(1)}\ \textup{Each member belongs to two and only two committees}\qquad \\ \text{(2)}\ \textup{Each pair of committees has one and only one member in common}$$

Then $x$: $\textbf{(A)} \ \textup{cannot be determined} \qquad \\ \textbf{(B)} \ \textup{has a single value between 8 and 16} \qquad \\ \textbf{(C)} \ \textup{has two values between 8 and 16} \qquad \\ \textbf{(D)} \ \textup{has a single value between 4 and 8} \qquad \\ \textbf{(E)} \ \textup{has two values between 4 and 8}$

## Solution

We can label each of the $x$ members with the pair of committees that they belong to, which is clearly valid due to rule (1). Then, by rule (2), for each pair of committees, there is exactly one member labeled with that pair. But since we have four committees, there must be $x = \binom{4}{2} = 6$ members in total. Thusly our choice is $\boxed{\textbf{(D)}}$, and we are done.

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