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how many subsets does {E,F,G,H} have










































How many subsets does {E,F,G,H} have?

Note: Each possible subset contains at least one of the elements the set.

For example, { E } and { F, H } are both subsets of {E,F,G,H}.

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Apr 02, 2012
Number of Subsets for {E,F,G,H}
by: Staff

Question:
how many subsets does {E,F,G,H} have?


Answer:

A subset contains at least one of the elements the set. For example, { E } and { F, H } are both subsets of {E,F,G,H}.

IMPROPER SUBSET:

An improper subset contains ALL the elements of the set. {E,F,G,H} is the improper subset of {E,F,G,H}.

PROPER SUBSET:

A proper subset contains one or more of the elements the set, but not all the elements. For example, { E } and { F, H } are proper subsets of {E,F,G,H}.


ELEMENTS in a set or subset CAN BE LISTED MORE THAN ONCE without changing the set or subset.

For example, { E, E, E} and { E, G, G } are still proper subsets of {E,F,G,H}.


ELEMENTS in a set or subset CAN BE LISTED IN ANY ORDER without changing the set or subset.

For example, {E,F,G,H} and {E, H, F,G, } are the same set, even though the elements are not listed in the same order.


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Subsets of {E,F,G,H}

The following calculations do not include the “null set” (the empty set), Ø = {}.



Improper Subset = 1: {E,F,G,H}

Proper Subsets = 30 (see calculations below):

To find the number of proper subsets, you must determine how many COMBINATIONS (not permutations) of the elements mom, dad, son, and daughter are possible when you select 1 element, 2 elements, or 3 elements (you cannot select all 4 elements because that would be an improper subset, which you have already accounted for).

There is a formula which you can use to calculate these combinations (again, not permutations, but combinations):

C(n,r) = n! / r! (n - r)!

0 ≤ r ≤ n.
n = number of elements in the set { E,F,G,H } = 4
r = number of elements selected = 1, 2, or 3, [r = 4 will not be used since that is the improper subset. It has already been accounted for.]

Order is not important

Repetition is not allowed


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Apr 02, 2012
Number of Subsets for {E,F,G,H}
by: Staff

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Part II



Selecting any 1 element from the 4 possible choices = 4 subsets possible
n = number of elements = 4
r = number of elements selected = 1

C(n,r) = n! / r! (n - r)!

C(4,1) = 4! / 1! (4 - 1)!
C(4,1) = (4*3*2*1) / 1*3*2*1
C(4,1) = 24 / 6
C(4,1) = 4

The proper subsets when 1 element is selected are: {E} {F} {G} {H}


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Selecting any 2 elements from the 4 possible choices = 6 subsets possible
n = number of elements = 4
r = number of elements selected = 2

C(n,r) = n! / r! (n - r)!


C(4,2) = 4! / 2! (4 - 2)!
C(4,2) = (4*3*2*1) / 2*1*2*1
C(4,2) = 12*2 / 2*2
C(4,2) = 24 / 4

C(4,2) = 6

The proper subsets when any 2 elements are selected: {E,F} {E,G} {E,H} {F,G} {F,H} {G,H}
-----------------------------------

Selecting any 3 elements from the 4 possible choices = 4 subsets possible
n = number of elements ( = 4)
r = number of elements selected (= 3)

C(n,r) = n! / r! (n - r)!

C(4,3) = 4! / 3! (4 - 3)!
C(4,3) = (4*3*2*1) /(3*2*1*1)
C(4,3) = 24 / 6
C(4,3) = 4

The proper subsets possible when any 3 elements are selected: {E,F,G} {E,F,H} {E,G,H} {F,G,H}

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>>>The FINAL ANSWER:

The number of all possible subsets (including the improper subset) of { E,F,G,H } = 18

(This DOES NOT include the “null set”, Ø = {}. If you wish to include the null set, the final answer is 15 + 1 = 16, or 2⁴. 2⁴ = 2ⁿ where n = 4 )

These are:

Proper Subsets:

{E} {F} {G} {H}

{E,F} {E,G} {E,H} {F,G} {F,H} {G,H}

{E,F,G} {E,F,H} {E,G,H} {F,G,H}


Improper Subset:

{ E,F,G,H }




Thanks for writing.

Staff
www.solving-math-problems.com


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