M28-30

Question

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

Bunuel · Answer

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

madhavmarda · Answer

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.

Bunuel · Answer

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.

kishgau · Answer

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!

luisnavarro · Answer

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

Bunuel · Answer

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}

luisnavarro · Answer

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

Bunuel · Answer

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.

luisnavarro · Answer

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

Bunuel · Answer

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}.

vijaisingh2001 · Answer

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

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

tukaisg · Answer

The formula 2^n for calculating the number of subsets comes from the fundamental principle of counting in combinatorics. Here's an explanation:
For each element in the set, we have two choices: include it in a subset or exclude it.
These choices are independent for each element. This means for each element, we multiply the number of choices.
With 5 elements, we have 2 choices for each, so we multiply 2 five times:
2 2 2 2 2 = 2^5 = 32
In general, for a set with n elements, we have 2 choices for each of the n elements, so the total number of subsets is 2^n.
This approach accounts for all possible combinations of including or excluding each element, which gives us all possible subsets, including the empty set and the full set itself.

Aryan19852 · Answer

Is my reasoning correct?
Except 0 we can select rest 5 single digit 5 ways, 2 in 5c2, 3 in 5c3, 4 in 5c4 and 5c5+ 0 subset which is 1
so we get 5+10+10+5+1+1=32

Rod728 · Answer

Bunuel,

I solved it correctly with 5C0+5C1+5C2+5C3+5C4+5C5=32. However, for other Subset questions, if they do not specifically mention "including the empty set", should I consider it as an option?

Thanks!

Bunuel · Answer

Yes, because an empty set is a subset of every set.

bumpbot · Answer

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M28-30

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