# List all the Subsets - College Math

List all the subsets of { 8, 15, 28, 41, 60}

identify the IMPROPER SUBSET (which contains all the elements of the original set).

identify all PROPER SUBSETs (sets which contain one or more of the elements of the original set)

ELEMENTS in the subsets CAN BE LISTED IN ANY ORDER.

### Comments for List all the Subsets - College Math

 Feb 26, 2012 List all the Subsets by: Staff Question:List all the subsets of { 8, 15, 28, 41, 60}Answer:A subset contains at least one of the elements the set. For example, { 8 } and { 15, 28 } are both subsets of { 8, 15, 28, 41, 60 }.IMPROPER SUBSET:An improper subset contains ALL the elements of the set. { 8, 15, 28, 41, 60 } is the improper subset of { 8, 15, 28, 41, 60 }.PROPER SUBSET:A proper subset contains one or more of the elements the set, but not all the elements. For example, { 8 } and { 15, 28 } are proper subsets of { 8, 15, 28, 41, 60 }.ELEMENTS in a set or subset CAN BE LISTED MORE THAN ONCE without changing the set or subset.For example, { 8, 8, 8} and { 15, 28, 28 } are still proper subsets of { 8, 15, 28, 41, 60 }.ELEMENTS in a set or subset CAN BE LISTED IN ANY ORDER without changing the set or subset.For example, { 8, 15, 28, 41, 60 } and { 28, 8, 15, 60, 41 } are the same set, even though the elements are not listed in the same order.------------------------------------Subsets of { 8, 15, 28, 41, 60 }The following calculations do not include the “null set” (the empty set), Ø = {}.Improper Subset = 1: { 8, 15, 28, 41, 60 }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 { 8, 15, 28, 41, 60 } = 5r = number of elements selected = 1, 2, 3, or 4 [r = 5 will not be used since that is the improper subset. It has already been accounted for.]Order is not importantRepetition is not allowed---------------------------------------

 Feb 26, 2012 List all the Subsets by: Staff ---------------------------------------Part IISelecting any 1 element from the 5 possible choices = 5 subsets possiblen = number of elements = 5r = number of elements selected = 1C(n,r) = n! / r! (n - r)!C(5,1) = 5! / 1! (5 - 1)!C(5,1) = (5*4*3*2*1) / 1*4*3*2*1C(5,1) = 5 / 1C(5,1) = 5The proper subsets when 1 element is selected are: {8} {15} {28} {41} {60 }-----------------------------------Selecting any 2 elements from the 5 possible choices = 10 subsets possiblen = number of elements = 5r = number of elements selected = 2C(n,r) = n! / r! (n - r)!C(5,2) = 5! / 2! (5 - 2)!C(5,2) = (5*4*3*2*1) / 2*1*3*2*1C(5,2) = 5*4 / 2*1C(5,2) = 20 / 2C(4,2) = 10The proper subsets when any 2 elements are selected: {8,15} {8,28} {8,41} {8,60} {15,28} {15,41} {15,60} {28,41} {28,60} {41,60}-----------------------------------Selecting any 3 elements from the 5 possible choices = 10 subsets possiblen = number of elements ( = 5)r = number of elements selected (= 3)C(n,r) = n! / r! (n - r)!C(4,3) = 5! / 3! (5 - 3)!C(4,3) = (5*4*3*2*1) /(3*2*1*2*1)C(4,3) = 20 / 2C(4,3) = 10The proper subsets possible when any 3 elements are selected: {8,15,28} {8,15,41} {8,15,60} {8,28,41} {8,28,60} {8,41,60} {15,28,41} {15,28,60} {15,41,60} {28,41,60}-----------------------------------Selecting any 4 elements from the 5 possible choices = 5 subsets possiblen = number of elements ( = 5)r = number of elements selected (= 4)C(n,r) = n! / r! (n - r)!C(4,3) = 5! / 4! (5 - 4)!C(4,3) = (5*4*3*2*1) /(4*3*2*1*1)C(4,3) = 5 / 1C(4,3) = 5The proper subsets possible when any 4 elements are selected: {8,15,28,41} {8,15,28,60} {8,15,41,60} {8,28,41,60} {15,28,41,60}-----------------------------------all the possible subsets of { 8, 15, 28, 41, 60}Improper Subsets = 1Proper Subsets = 5 + 10 + 10 + 5 = 30>>>The FINAL ANSWER:The number of all possible subsets (including the improper subset) of { 8, 15, 28, 41, 60} = 31(This DOES NOT include the “null set”, Ø = {}. If you wish to include the null set, the final answer is 31 + 1 = 32, or 2⁵. 2⁵ = 2ⁿ where n = 5 )These are: Proper Subsets:{8} {15} {28} {41} {60 }{8,15} {8,28} {8,41} {8,60} {15,28} {15,41} {15,60} {28,41} {28,60} {41,60}{8,15,28} {8,15,41} {8,15,60} {8,28,41} {8,28,60} {8,41,60} {15,28,41} {15,28,60} {15,41,60} {28,41,60}{8,15,28,41} {8,15,28,60} {8,15,41,60} {8,28,41,60} {15,28,41,60}Improper Subset:{ 8, 15, 28, 41, 60} Thanks for writing.Staff www.solving-math-problems.com

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