An algebra problem by Anish Roy

We consider the six subsets of possible combinations as: $A_{1}={(x,17,24):1 \leq x \leq 40}$ $A_{2}={(x,24,17):1 \leq x \leq 40}$ $A_{3}={(17,x,24):1 \leq x \leq 40}$ $A_{4}={(24,x,17):1 \leq x \leq 40}$ $A_{5}={(17,24,x):1 \leq x \leq 40}$ $A_{6}={(24,17,x):1 \leq x \leq 40}$ It is not difficult to see that each set has $40$ elements and so by addition principle ther are $40\times6=240$ combinations to try and at most $40$ minutes are needed. An important but easily missed fact in applying the addition principle is that the sets $A_{i}$ must form a partition for the principle to hold; that is, $A_{i} \cap A_{j}=\phi$ for $i \neq j$ . But in this problem, the combinations $(17,17,24)$ belongs to both $A_{1}$ and $A_{3}$ . Similarly, each of the combinations $(17,24,17),(17,24,24),(24,17,24),(24,24,17),(24,17,17)$ also belongs to two sets and is therefore counted twice. Hence, there are only $240-6=234$ combinations to try, and the correct answer is $\boxed{39}$ minutes.