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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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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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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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In a certain mathematical activity, we have five cards with five diffe [#permalink]
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17 Mar 2015, 19:03
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Case 1: 1 used envelope => 1 way Case 2: 2 used envelopes  410: Choose 4 from 5 cards: 5 ways  320: Choose 3 from 5 cards: 10 ways Case 3: All envelopes used  311: Choose 3 from 5 and no need to choose 1 from 2: 10 ways  221: Choose 2 from 5 and choose 2 from 3, but two groups are the same => (10X3):2 = 15
Total: 1+5+10+10+15=41 => Answer: A
Last edited by camlan1990 on 17 Mar 2015, 19:27, edited 3 times in total.



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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
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camlan1990 wrote: Case 1: 1 used envelope => 1 way Case 2: 2 used envelopes  410: Choose 4 from 5 cards: 5 ways  320: Choose 3 from 5 cards: 10 ways Case 3: All envelopes used  311: Choose 3 from 5 and choose 1 from 2: 10x2 = 20 ways  221: Choose 2 from 5 and choose 2 from 3, but two groups are the same => (10X3):2 = 15
Total: 1+5+10+20+15=41 => Answer: A hi your total is 51 and not 41.. i am sure you would want to half the ways in case 3 for 311.. 20/2=10 ways as you have done for 221... you will then get 41... but just a question, why are you taking the two groups as same.. they may have 22 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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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 311: 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
chetan2uHi chetan Can you please help me in understanding this question as i am not able to grab it properly Regards SG



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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
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: 311 split = 5C3 = 10 possibilities Case V: 221 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 doublecounts the two pairs, so we need to divide by two. 15 possibilities. N = 1 + 5 + 10 + 10 + 15 = 41 Answer = (A)
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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: 311 split = 5C3 = 10 possibilities Case V: 221 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 doublecounts the two pairs, so we need to divide by two. 15 possibilities. N = 1 + 5 + 10 + 10 + 15 = 41 Answer = (A) "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: 311 split = 5C3 = 10 possibilities Case V: 221 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 doublecounts the two pairs, so we need to divide by two. 15 possibilities. N = 1 + 5 + 10 + 10 + 15 = 41 Answer = (A) "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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12 Apr 2017, 11:31
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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) Possibility 1a: Envelop1 receive 0card  Envelop2 receive 0card  Envelop3 receive 5cards === 112310 Possibility 1b: Envelop1 receive 0card  Envelop2 receive 5card  Envelop3 receive 0card === 123101 Possibility 1c: Envelop1 receive 5cards  Envelop2 receive 0card  Envelop3 receive 0card === 231011 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 410 is not an acceptable set because any prime number is greater than 1. same logic, case 140 is not an acceptable set. case 014 and case 041 toghether have 5 acceptable sets. Possibilites that are not accpetable in case 014 are acceptable in case 041 and visversa. Same logic can be replicate to Case 023 and Case 032, which together have 10 acceptable sets. Same for case 113, which has 10 acceptable sets. The trick is then on case 122, 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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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 splitups for 5 Prime Numbers are as follows: 500: No. Of Sets possible: 5C5 = 1 410: No. Of Sets possible: 5C4 x 1C1 = 5 x 1= 5 320: No. Of Sets possible: 5C3 x 2C2 = 10 x 1 = 10 311: No. Of Sets possible: {5C3 x 2C1 x 1C1} / 2! ( 2! Is the correction for identical 1's)= 10 221: 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. Hence Answer Choice : A




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