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# In a certain mathematical activity, we have five cards with five diffe

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Math Expert
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In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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17 Mar 2015, 04:42
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95% (hard)

Question Stats:

21% (02:56) correct 79% (02:38) wrong based on 221 sessions

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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

Kudos for a correct solution.

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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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23 Mar 2015, 03:16
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Bunuel wrote:
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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

We have to consider different groupings. First, all the cards together.

Case I: all five cards in one envelope (2 envelopes empty) = one possibility

Now, two envelopes used, one empty.

Case II: four in one, one in another, one empty = five choices for the one by itself = 5 possibilities

Case III: three in one, two in another, one empty = 5C3 = 10 possibilities

Now, no empty envelopes, all three used:

Case IV: 3-1-1 split = 5C3 = 10 possibilities

Case V: 2-2-1 split = 5C2 = 10 for the first pair, then 3 possibilities for the single one, leaving the other two for the pair. This would be 10*3 = 30, but that double-counts the two pairs, so we need to divide by two. 15 possibilities.

N = 1 + 5 + 10 + 10 + 15 = 41

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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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17 Apr 2017, 15:40
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There are 5 Different Prime Numbers that needs to be split among three identical envelopes.
Possible split-ups for 5 Prime Numbers are as follows:
5-0-0: No. Of Sets possible: 5C5 = 1
4-1-0: No. Of Sets possible: 5C4 x 1C1 = 5 x 1= 5
3-2-0: No. Of Sets possible: 5C3 x 2C2 = 10 x 1 = 10
3-1-1: No. Of Sets possible: {5C3 x 2C1 x 1C1} / 2! ( 2! Is the correction for identical 1's)= 10
2-2-1: No. Of Sets possible: {5C2 x 3C2 x 1C1} / 2! ( 2! Is the correction for identical 2's)= 15

Counting total sets possible= 1 + 5 + 10 + 10 + 15 = 41.
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In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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17 Mar 2015, 19:21
2
camlan1990 wrote:
Case 1: 1 used envelope => 1 way
Case 2: 2 used envelopes
- 4-1-0: Choose 4 from 5 cards: 5 ways
- 3-2-0: Choose 3 from 5 cards: 10 ways
Case 3: All envelopes used
- 3-1-1: Choose 3 from 5 and choose 1 from 2: 10x2 = 20 ways
- 2-2-1: Choose 2 from 5 and choose 2 from 3, but two groups are the same => (10X3):2 = 15

hi your total is 51 and not 41..
i am sure you would want to half the ways in case 3 for 3-1-1.. 20/2=10 ways as you have done for 2-2-1...
you will then get 41...
but just a question, why are you taking the two groups as same..
they may have 2-2 prime numbers in each but remember the product would be different for each envelope everytime as no two different prime numbers can give you same answer when multiplied..
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In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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Updated on: 17 Mar 2015, 19:27
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1
Case 1: 1 used envelope => 1 way
Case 2: 2 used envelopes
- 4-1-0: Choose 4 from 5 cards: 5 ways
- 3-2-0: Choose 3 from 5 cards: 10 ways
Case 3: All envelopes used
- 3-1-1: Choose 3 from 5 and no need to choose 1 from 2: 10 ways
- 2-2-1: Choose 2 from 5 and choose 2 from 3, but two groups are the same => (10X3):2 = 15

Originally posted by camlan1990 on 17 Mar 2015, 19:03.
Last edited by camlan1990 on 17 Mar 2015, 19:27, edited 3 times in total.
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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12 Apr 2017, 11:31
1
This is a very tough one!

Product of the 5 different numbers 2*3*5*7*11=2310 (Prime numbers 2,3,5,7 and 11 are chosen just for example)

The 3 possiblities above will result in 1 set because we must put the envelop in order (lowest to highest).
in other words, we eliminate possibilities in which the numbers are not in increasing order.

So, solve this problem, you can asume that the first envelop has the lowest product and the third envelop has the highest product.

Similar logic tells you that there are 5 sets possible in the case: envelop1 has 0 card, envelop2 has 1 card and envelop3 has 4 cards
in fact, case 4-1-0 is not an acceptable set because any prime number is greater than 1. same logic, case 1-4-0 is not an acceptable set.
case 0-1-4 and case 0-4-1 toghether have 5 acceptable sets. Possibilites that are not accpetable in case 0-1-4 are acceptable in case 0-4-1 and vis-versa.

Same logic can be replicate to Case 0-2-3 and Case 0-3-2, which together have 10 acceptable sets.
Same for case 1-1-3, which has 10 acceptable sets.

The trick is then on case 1-2-2, in which half of the possibilities are not in ascending order and are not replaced in any other case.
So from the 30 possibilities (5C1*2C4), only 15 are in ascending order, therefore acceptable sets.

The total acceptable cases are 1+5+10+10+15, ie 41 sets
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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24 Jun 2018, 03:50
1
TheSpecialNone wrote:
Would it make any difference if the products of the cards exceed the number of cards ?
Ex: we have the first 5 prime numbers - 2,3,5,7 and 11- and only 2 envelopes used
For example 3-2-0 and 2,3,5 and 7,11 respectively inside, where the first product is smaller than the second.
Does it have any consequences in the problem ?

Yes exactly. They have not considered the case when product is less than a particular prime number.
For example if prime numbers are 2,3,5,7,101.
This kind of case has not been considered.
And even the solution given by Magoosh official sums to 51 not 41.
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In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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17 Mar 2015, 17:34
I believe the answer is B.

Each of the five cards can go in 1 of 3 envelopes, so there are 3^5 possible combinations.
As we are using different prime numbers, the possible products will always be distinct from each other and there will be no overlap.
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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17 Mar 2015, 19:30
Hi chetan2u,

I m sorry for this mistake. I highlighted my answer with RED above.

In case 3-1-1: After choosing 3 from 5, you dont need to choose 1 from 2. Because there is ONE way to put in order for two other cards.
Thanks
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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18 Mar 2015, 11:49
chetan2u

Hi chetan
Can you please help me in understanding this question as i am not able to grab it properly

Regards
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In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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19 Jun 2015, 03:54
Bunuel wrote:
Bunuel wrote:
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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

We have to consider different groupings. First, all the cards together.

Case I: all five cards in one envelope (2 envelopes empty) = one possibility

Now, two envelopes used, one empty.

Case II: four in one, one in another, one empty = five choices for the one by itself = 5 possibilities

Case III: three in one, two in another, one empty = 5C3 = 10 possibilities

Now, no empty envelopes, all three used:

Case IV: 3-1-1 split = 5C3 = 10 possibilities

Case V: 2-2-1 split = 5C2 = 10 for the first pair, then 3 possibilities for the single one, leaving the other two for the pair. This would be 10*3 = 30, but that double-counts the two pairs, so we need to divide by two. 15 possibilities.

N = 1 + 5 + 10 + 10 + 15 = 41

"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.". I could not understand these statements and how we used these statements in the solution. Very confusing..please help..
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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19 Jun 2015, 07:04
VerticalCharlie wrote:
Bunuel wrote:
Bunuel wrote:
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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

We have to consider different groupings. First, all the cards together.

Case I: all five cards in one envelope (2 envelopes empty) = one possibility

Now, two envelopes used, one empty.

Case II: four in one, one in another, one empty = five choices for the one by itself = 5 possibilities

Case III: three in one, two in another, one empty = 5C3 = 10 possibilities

Now, no empty envelopes, all three used:

Case IV: 3-1-1 split = 5C3 = 10 possibilities

Case V: 2-2-1 split = 5C2 = 10 for the first pair, then 3 possibilities for the single one, leaving the other two for the pair. This would be 10*3 = 30, but that double-counts the two pairs, so we need to divide by two. 15 possibilities.

N = 1 + 5 + 10 + 10 + 15 = 41

"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.". I could not understand these statements and how we used these statements in the solution. Very confusing..please help..

Say the 5 primes are: 2, 3, 5, 7, and 11.

These primes then are distributed in 3 envelopes. Say 2 and 3 are in envelope #1, 5 and 7 envelope #2 and 11 in envelope #3.

The product of primes in envelopes are ("numbers" of envelopes):
For #1: 2*3 = 6;
For #2: 5*7 = 35;
For #3: 11.

So, set we get with this distribution is {6, 11, 35}. This is one of the sets possible.

The question asks how many different sets can be produced?

Hope it's clear.
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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12 Apr 2017, 08:26
I dont quite get it, what the prime number has to do with the question? I understand that those prime numbers should be factored into the solution.
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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26 Apr 2018, 18:24
camlan1990 wrote:
Case 1: 1 used envelope => 1 way
Case 2: 2 used envelopes
- 4-1-0: Choose 4 from 5 cards: 5 ways
- 3-2-0: Choose 3 from 5 cards: 10 ways
Case 3: All envelopes used
- 3-1-1: Choose 3 from 5 and no need to choose 1 from 2: 10 ways
- 2-2-1: Choose 2 from 5 and choose 2 from 3, but two groups are the same => (10X3):2 = 15

Is there a reason we don't need to choose 1 from 2
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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29 May 2018, 06:34
Would it make any difference if the products of the cards exceed the number of cards ?
Ex: we have the first 5 prime numbers - 2,3,5,7 and 11- and only 2 envelopes used
For example 3-2-0 and 2,3,5 and 7,11 respectively inside, where the first product is smaller than the second.
Does it have any consequences in the problem ?
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Re: In a certain mathematical activity, we have five cards with five diffe  [#permalink]

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Re: In a certain mathematical activity, we have five cards with five diffe   [#permalink] 19 Jul 2019, 11:56
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