**Question:**

In a certain mathematical activity, we have five cards with five different prime numbers on them. We will distribute these five cards among three envelope: all could go in any envelope, or they could be broken up in any way among the envelopes. Then in each envelop, we find the product of all the cards in that envelope: that is the “number” of the envelope. An envelope containing no cards has the number 1. We then put the three envelope numbers in order, from lowest to highest, and that is our set. How many different sets can be produced by this process?

A. 41

B. 89

C. 125

D. 243

E. 512

**MBA Wisdom's Answer:**

$$\text{Let's consider the different cases:}$$ $$\text{Case 1: 5-0-0}$$ $$\text{Possibilities} = \text{5C5} = {1}$$ $$\text{Case 2: 4-1-0}$$ $$\text{Possibilities} = \text{5C4} = {5}$$ $$\text{Case 3: 3-2-0}$$ $$\text{Possibilities} = \text{5C3} = {10}$$ $$\text{Case 4: 3-1-1}$$ $$\text{Possibilities} = \text{5C3} = {10}$$ $$\text{Case 5: 2-2-1}$$ $$\text{Possibilities} = {{\text{5C2} \times \text{3C2}} \over {2!}} = {{\left({10}\right) \left({3}\right)} \over {2}} = {15}$$ $$$$ $$\text{Total numbet of sets:} = {1} + {5} + {10} + {10} + {15} = {41}$$

**A. 41**