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M28-30

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Bunuel
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M28-30 [#permalink] New post   16 Sep 2014, 01:29
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How many different subsets, including the empty set, of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. \(16\)
B. \(27\)
C. \(31\)
D. \(32\)
E. \(64\)
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D
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Re M28-30 [#permalink] New post   16 Sep 2014, 01:29
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Official Solution:

How many different subsets, including the empty set, 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 of the 5 elements in the set {1, 2, 3, 4, 5} has two possibilities: either to be included in a subset or to be excluded. Therefore, the total number of subsets for this set is \(2^5 = 32\). These 32 subsets include the empty set, single-element subsets like {1}, {2}, {3}, {4}, and {5}, and multi-element subsets like {1, 2}, {1, 2, 3}, {1, 3, 4, 5}, and so on. Each of these subsets will also be a subset of {0, 1, 2, 3, 4, 5} and will not contain the element 0.


Answer: D
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Re: M28-30 [#permalink] New post   24 Sep 2014, 23:30
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: M28-30 [#permalink] New post   25 Sep 2014, 02:04
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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 non-empty set, so no, we don't have to subtract 1.
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Re: M28-30 [#permalink] New post   23 Oct 2014, 05:28
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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: M28-30 [#permalink] New post   03 Jul 2015, 05:20
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.

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Luis Navarro
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Re: M28-30 [#permalink] New post   03 Jul 2015, 06:12
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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: M28-30 [#permalink] New post   03 Jul 2015, 12:38
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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Bunuel
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Re: M28-30 [#permalink] New post   05 Jul 2015, 08:31
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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 5-element sets but arranged in different ways, while subsets give an empty set, 1-element sets, 2-elements set, ..., 5-element sets.

Hope it's clear.
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Re: M28-30 [#permalink] New post   05 Jul 2015, 09:03
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 5-element sets but arranged in different ways, while subsets give an empty set, 1-element sets, 2-elements set, ..., 5-element 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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Bunuel
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Re: M28-30 [#permalink] New post   05 Jul 2015, 09:06
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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 5-element sets but arranged in different ways, while subsets give an empty set, 1-element sets, 2-elements set, ..., 5-element 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: M28-30 [#permalink] New post   09 Sep 2016, 13:38
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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: M28-30 [#permalink] New post   20 Jun 2023, 07:31
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M28-30 [#permalink]
20 Jun 2023, 07:31
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