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Re M2830 [#permalink]
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16 Sep 2014, 00:29
Official Solution:How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0? A. \(16\) B. \(27\) C. \(31\) D. \(32\) E. \(64\) Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is \(2^5=32\). Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0. Answer: D
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Re: M2830 [#permalink]
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24 Sep 2014, 22:30
Hi Bunuel,
In this question should we not subtract 1 from 2^5 to get the answer as 31. Because that one scenario would mean that none of 1,2,3,4,5 are included in the particular subset. Please advise. Thank You in advance.



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Re: M2830 [#permalink]
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25 Sep 2014, 01:04
madhavmarda wrote: Hi Bunuel,
In this question should we not subtract 1 from 2^5 to get the answer as 31. Because that one scenario would mean that none of 1,2,3,4,5 are included in the particular subset. Please advise. Thank You in advance. Set which does not contain any of the terms is an empty set. An empty set is a subset of every nonempty set, so no, we don't have to subtract 1.
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Re: M2830 [#permalink]
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23 Oct 2014, 04:28
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From {1,2,3,4,5}, the sets could be formed as 5C1 + 5C2 + 5C3 + 5C4 + 5C5 + 5C0 ( the empty set). I see what you mean. Thank you!



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Re: M2830 [#permalink]
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03 Jul 2015, 04:20
Bunuel wrote: Official Solution:
How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. \(16\) B. \(27\) C. \(31\) D. \(32\) E. \(64\)
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is \(2^5=32\). Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D Hi, I find it hard to understand why it should not be 5¡, because it is suppossed that there are 5 choices that you are combining; I think that answer 2^5=32 seems logic but It is hard to me to visualize why 5¡ does not work here. Thanks a lot. Regards Luis Navarro Looking for 700



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Re: M2830 [#permalink]
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03 Jul 2015, 05:12
luisnavarro wrote: Bunuel wrote: Official Solution:
How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. \(16\) B. \(27\) C. \(31\) D. \(32\) E. \(64\)
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is \(2^5=32\). Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D Hi, I find it hard to understand why it should not be 5¡, because it is suppossed that there are 5 choices that you are combining; I think that answer 2^5=32 seems logic but It is hard to me to visualize why 5¡ does not work here. Thanks a lot. Regards Luis Navarro Looking for 700 5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking. We need subsets without 0, which are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5}
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Re: M2830 [#permalink]
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03 Jul 2015, 11:38
Bunuel wrote: luisnavarro wrote: Bunuel wrote: Official Solution:
How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. \(16\) B. \(27\) C. \(31\) D. \(32\) E. \(64\)
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is \(2^5=32\). Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D Hi, I find it hard to understand why it should not be 5¡, because it is suppossed that there are 5 choices that you are combining; I think that answer 2^5=32 seems logic but It is hard to me to visualize why 5¡ does not work here. Thanks a lot. Regards Luis Navarro Looking for 700 5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking. We need subsets without 0, which are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5} Hi Bunuel, Sorry, I am a little confused yet, 5! gives different arrangements without 0 {1, 2, 3, 4, 5}, and in your explanation you said: 5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking. But I was reffering to 5! without cero... I am confused, could you help me again? Thanks a lot Regards Luis Navarro Looking for 700



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Re: M2830 [#permalink]
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05 Jul 2015, 07:31
luisnavarro wrote: Hi Bunuel,
Sorry, I am a little confused yet, 5! gives different arrangements without 0 {1, 2, 3, 4, 5}, and in your explanation you said:
5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking.
But I was reffering to 5! without cero... I am confused, could you help me again?
Thanks a lot
Regards
Luis Navarro Looking for 700 Yes, 5! gives different arrangements of {1, 2, 3, 4, 5}. 0 in my previous post was a typo. We need SUBSETS of {0, 1, 2, 3, 4, 5} without 0, not arrangements of {1, 2, 3, 4, 5}. Subsets of {0, 1, 2, 3, 4, 5} without 0 are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5} Notice that arrangements of {1, 2, 3, 4, 5} give the same 5element sets but arranged in different ways, while subsets give an empty set, 1element sets, 2elements set, ..., 5element sets. Hope it's clear.
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Re: M2830 [#permalink]
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05 Jul 2015, 08:03
Bunuel wrote: luisnavarro wrote: Hi Bunuel,
Sorry, I am a little confused yet, 5! gives different arrangements without 0 {1, 2, 3, 4, 5}, and in your explanation you said:
5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking.
But I was reffering to 5! without cero... I am confused, could you help me again?
Thanks a lot
Regards
Luis Navarro Looking for 700 Yes, 5! gives different arrangements of {1, 2, 3, 4, 5}. 0 in my previous post was a typo. We need SUBSETS of {0, 1, 2, 3, 4, 5} without 0, not arrangements of {1, 2, 3, 4, 5}. Subsets of {0, 1, 2, 3, 4, 5} without 0 are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5} Notice that arrangements of {1, 2, 3, 4, 5} give the same 5element sets but arranged in different ways, while subsets give an empty set, 1element sets, 2elements set, ..., 5element sets. Hope it's clear. Thanks, it is more clear know, I only have a little more doubt... What I inffer for this is that subsets orders by itself "automaticly", so then {1, 2, 3, 4, 5} only consider 1 posible combination, instead of mixing the order in the positions for instance: {3, 1, 5, 4, 2}, {2, 5, 3, 1, 4}... is my conclusion wrong or correct? Thanks a lot. Regards. Luis Navarro Looking for 700



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Re: M2830 [#permalink]
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05 Jul 2015, 08:06
luisnavarro wrote: Bunuel wrote: luisnavarro wrote: Hi Bunuel,
Sorry, I am a little confused yet, 5! gives different arrangements without 0 {1, 2, 3, 4, 5}, and in your explanation you said:
5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking.
But I was reffering to 5! without cero... I am confused, could you help me again?
Thanks a lot
Regards
Luis Navarro Looking for 700 Yes, 5! gives different arrangements of {1, 2, 3, 4, 5}. 0 in my previous post was a typo. We need SUBSETS of {0, 1, 2, 3, 4, 5} without 0, not arrangements of {1, 2, 3, 4, 5}. Subsets of {0, 1, 2, 3, 4, 5} without 0 are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5} Notice that arrangements of {1, 2, 3, 4, 5} give the same 5element sets but arranged in different ways, while subsets give an empty set, 1element sets, 2elements set, ..., 5element sets. Hope it's clear. Thanks, it is more clear know, I only have a little more doubt... What I inffer for this is that subsets orders by itself "automaticly", so then {1, 2, 3, 4, 5} only consider 1 posible combination, instead of mixing the order in the positions for instance: {3, 1, 5, 4, 2}, {2, 5, 3, 1, 4}... is my conclusion wrong or correct? Thanks a lot. Regards. Luis Navarro Looking for 700 A set, by definition, is a collection of elements without any order. ( While, a sequence, by definition, is an ordered list of terms.) So, we are not interested in ordering the elements in a set (subset): {1, 2} is the same set as {2, 1}.
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Re: M2830 [#permalink]
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05 Jul 2015, 11:12
Hi Bunuel,
I appreciatte your help, thanks a lot, it is totally clear now.
Best regards.
Luis Navarro Looking for 700



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Re: M2830 [#permalink]
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27 Sep 2015, 10:40
Bunuel wrote: madhavmarda wrote: Hi Bunuel,
In this question should we not subtract 1 from 2^5 to get the answer as 31. Because that one scenario would mean that none of 1,2,3,4,5 are included in the particular subset. Please advise. Thank You in advance. Set which does not contain any of the terms is an empty set. An empty set is a subset of every nonempty set, so no, we don't have to subtract 1. Hi, Bunuel. Could you please give a link to the rule about empty set being a subset. It sounds strange, taking into account that this term comes from statistics, which never studies 0 resptesentative set. TIA



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Re: M2830 [#permalink]
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27 Nov 2015, 11:37
kishgau wrote: From {1,2,3,4,5}, the sets could be formed as 5C1 + 5C2 + 5C3 + 5C4 + 5C5 + 5C0 ( the empty set). I see what you mean. Thank you! How does this work? 5C1 = 5 5C2 = 10 5C3 = 10 5C4 = 5 There are 30 subsets according to the above. Why would we include 5C5? 5C5 is the entire set, so by definition it is not a subset. Also, why would an empty set be considered a subset?



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Re: M2830 [#permalink]
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09 Sep 2016, 12:38
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seems answer choice D is wrong subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0? subset with just one number  5c1= 5  (1), (2), (3), (4), (5) subset with 02 numbers from(1,2,3,4,5)  5c2 = 10  example (1,2),(1,3)..... subset with 01 numbers  5c3= 10 example (1,2,3) (1,2,4) subset with 04 numbers 5c4= 5  example (1,2,3,4), (12,3,5)...... example with 05 numbers 5c5 = 1 the whole one set ( 1,2,3,4,5)
so total = 5+10+10+10+1= 31



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Re: M2830 [#permalink]
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17 Sep 2016, 00:48
Can anyone, please, explain what an empty subset is?



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Re: M2830 [#permalink]
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17 Sep 2016, 00:54
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vijaisingh2001 wrote: seems answer choice D is wrong subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0? subset with just one number  5c1= 5  (1), (2), (3), (4), (5) subset with 02 numbers from(1,2,3,4,5)  5c2 = 10  example (1,2),(1,3)..... subset with 01 numbers  5c3= 10 example (1,2,3) (1,2,4) subset with 04 numbers 5c4= 5  example (1,2,3,4), (12,3,5)...... example with 05 numbers 5c5 = 1 the whole one set ( 1,2,3,4,5)
so total = 5+10+10+10+1= 31 Empty subset needs to be considered Hence 32 Sent from my iPhone



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Re: M2830 [#permalink]
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17 Sep 2016, 00:55
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Devbek wrote: Can anyone, please, explain what an empty subset is? Happy to help The empty set is the set containing no elements. In mathematics, and more specifically set theory, theempty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero. Sent from my iPhone



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Re: M2830 [#permalink]
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17 Sep 2016, 01:39
paidlukkha wrote: Devbek wrote: Can anyone, please, explain what an empty subset is? Happy to help The empty set is the set containing no elements. In mathematics, and more specifically set theory, theempty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero. Sent from my iPhone Thanks a lot. Math still amazes me with its weird stuff



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Re M2830 [#permalink]
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05 Jun 2017, 05:25
The other solution to this could be : Number of ways a subset with only one number can be formed : 5 ways ( taking 1, 2, 3, 4 and 5 each) Number of ways a subset with two numbers can be formed : 5C2 = 10 ways Number of ways a subset with three numbers can be formed : 5C3 = 10 ways Number of ways a subset with four numbers can be formed : 5C4 = 1 way Null set( empty set) = 1 way
Answer: 5+10+10+5+1+1 =32 Things to remember while choosing : For example subset (1,2,3) is same as subset (3,2,1) .







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