Iranian Mathematical Olympiad

Let $a_1, a_2, \dots, a_n$ denote the lamps, and $b_1, b_2, \dots, b_5$ denote the keys. First, it is easy to see that $n \ge 3$, because otherwise there are at most 4 subsets of lamps and it's in contradiction with the assumption of the problem. Construct a $10 \times n$ table that each row corresponds to a triple set of keys and each column corresponds to a lamp.

In intersection of row $i$ and column $j$ assign number 1 if after switching 3 keys corresponding to row $i$, lamp $a_j$ turns on, and else assign 0. Denote the sum of row $i$'s numbers by $z_i$. Suppose lamp $a_i$ is connected to key $X_j$. Now sum of numbers in column $j$ is $\binom{X_j}{3} + \binom{X_j}{1}\binom{5-X_j}{2}$, so

$$\sum_{i=1}^{10} z_i = \sum_{j=1}^{n} \left( \binom{X_j}{3} + \binom{X_j}{1} \binom{5-X_j}{2} \right) \qquad (1)$$

$X_j$ is positive integer between 1 and 5. It's easy to check that $\binom{X_j}{3} + \binom{X_j}{1}\binom{5-X_j}{2}$ phrase is always greater than 3. Because of equation 1, it is deduced that

$$\sum_{i=1}^{10} z_i \geq 4n.$$

Because of pigeonhole principle there exist $i$ so that $z_i \geq \lfloor \frac{4n}{10} \rfloor \geq 2$ and we're done. ■