A restaurant has 5 desserts, 3 side dishes and 4 main dishes. A student chooses one side dish, one main dish, and one dessert. How many different meals could he make?

30
Explanation
if the first event occurs in x ways, and the second event occurs in y ways, then two events occur in as sequence of xy ways.
so
event A ; choose (1) dessert , 5 ways
event B , chosen (1) side dish, 3 ways
event C, choose (1) main dish, 2 ways
so
a meal( 1 dessert+1 side dish+main dish) is the product of the 3 ways
[tex]\begin{gathered} \text{ways a meal could be made= (5}\cdot3\cdot2)\text{ ways} \\ \text{ways a meal could be made=}30\text{ ways} \end{gathered}[/tex]therefore, the answer is
30
I hope this helps you