# 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}.

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

 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. ------------------------------------ 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 ---------------------------------------

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