In a certain mathematical activity, ww start with seven...

This questions looks much more complicated than it really is. Because we only have prime numbers, and there is no way to multiply different prime numbers together to get the same product, all of the possible "numbers" we can get for each box will be different.

So for the 4 card box, we pick 4 out of 7, where order doesn't matter:

$$\frac{7!}{\left(7-4\right)!4!}=35$$ So we have 35 possibilities for the first box.

Then for the 2 card box, we pick 2 out of the remaining 3, where order doesn't matter:

$$\frac{3!}{\left(3-2\right)!2!}=\ 3$$ So we have 3 possibilities for the second box, once the first box has been picked.

Then we have one remaining choice for our 1 card box. So we have 1 possibility for the third box, once the first and second box have been picked.

This means that in total, we have 35*3*1=105 possible combinations for our three boxes.

Now, once we have our combination of 3 products, there is only 1 way for us to arrange them — from lowest to highest. So for each of the 105 possible combinations, there is only one possible set. This means that there are 105 possible sets, or answer choice B.