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

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In a certain mathematical activity, we start with seven cards, each with a different prime number written on it. These seven cards are randomly put into three boxes in the following way: one box must get four cards, one must get two, and one gets only one. Then, for each box, we find the product of all the cards in the box, and that's the "number" of the box. Then, we put those three numbers in order, from lowest to highest, and that is our set. How many different sets can be created from this process?

A. 35
B. 105
C. 210
D. 420
E. 630

The OA is B.

I'm really confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance.

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by ErikaPrepScholar » Tue Mar 20, 2018 4:44 am
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.
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