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# In a certain mathematical activity, we start with seven cards, each wi

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Joined: 02 Sep 2009
Posts: 58401

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17 Mar 2015, 05:43
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65% (hard)

Question Stats:

59% (02:26) correct 41% (02:54) wrong based on 74 sessions

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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

Kudos for a correct solution.

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17 Mar 2015, 06:28
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Bunuel wrote:
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

Kudos for a correct solution.

7C4 * 3C2 = 105 Answer B.

7C4 ways to choose 4 numbers for Box 1
3C2 ways to choose 2 numbers for Box2
1 way for Box 3 .
every combination will give a different product and can be arranged least to highest only in one way .
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17 Mar 2015, 06:42
3
It is important to note that since we're dealing with primes, the product will be different. Also, the last part of ordering from greatest to least has no impact.

7!/(4! 3!) × 3!/(2! 1!) × 1 = 105

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18 Mar 2015, 11:16
1
Bunuel wrote:
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

Kudos for a correct solution.

7! / 4! 2! 1!
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Joined: 02 Sep 2009
Posts: 58401

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23 Mar 2015, 04:14
1
Bunuel wrote:
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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Part of the logic of the problem involves recognizing that every unique combination of prime numbers produces a unique product. There is no way that two different groupings cards will produce the same set of numbers.

First of all, the cards to go into the cup holding 4 cards.

7C4 = 35.

Once we have placed 4 card in the first cup, we have three cards left, which means we have three choices of a single card to put into the third cup. Three choices. Once we place that, the remaining two cards must go into the second cup: no choice there.

N = 35*3 = 105

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10 Apr 2016, 14:01
7C4 *3C2 = 35 *3 = 105
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10 Apr 2019, 07:45
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In a certain mathematical activity, we start with seven cards, each wi   [#permalink] 10 Apr 2019, 07:45
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