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

[BTGmoderatorLU]

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 fours 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 with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

[DavidG@VeritasPrep]

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 fours 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 with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

Box #1 (four cards) We've got 7 numbers to select from. We want to select 4 of them. So we want 7C4 = 7*6*5*4/4*3*2*1 = 35 options for the first box.

Box #2 (two cards). If we started with 7 numbers, and we've selected 4, we only have 3 cards remaining. Of these 3, we want to select 2, so we want 3C2 = 3*2/2*1 = 3 options for the second box.

Box #3 (one card). We started with 7 numbers. We selected 4 for the first box and 2 for the second box, so there's only one number left. That would leave us 1 lonely option for the last box.

Multiply the results: 35 * 3 * 1 = 105. The answer is B