# how many subsets in P={c,p,l,a}

by Kim
(Riv. CA)

How many subsets in P={c,p,l,a}?

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

For example, { c,p,l,a } and {p,l,a,c} are the same set, even though the elements are not listed in the same order.

### Comments for how many subsets in P={c,p,l,a}

 Apr 12, 2012 Number of Subsets in {c,p,l,a} by: Staff Question: by Kim (Riv. CA) how many subsets in P={c,p,l,a} Answer: A subset contains at least one of the elements the set. For example, { c } and { p, a } are both subsets of {c,p,l,a}. IMPROPER SUBSET: An improper subset contains ALL the elements of the set. { c,p,l,a } is the improper subset of { c,p,l,a }. PROPER SUBSET: A proper subset contains one or more of the elements the set, but not all the elements. For example, { c } and { p,a } are proper subsets of { c,p,l,a }. ELEMENTS in a set or subset CAN BE LISTED MORE THAN ONCE without changing the set or subset. For example, { a,a,a} and { l,c,c } are still proper subsets of { c,p,l,a }. ELEMENTS in a set or subset CAN BE LISTED IN ANY ORDER without changing the set or subset. For example, { c,p,l,a } and {p,l,a,c} are the same set, even though the elements are not listed in the same order. ------------------------------------ Subsets of { c,p,l,a } The following calculations do not include the “null set” (the empty set), Ø = {}. Improper Subset = 1: { c,p,l,a } Proper Subsets = (see calculations below): To find the number of proper subsets, you must determine how many COMBINATIONS (not permutations) of the elements c, p, l, a 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 { c,p,l,a } = 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 12, 2012 Number of Subsets in {c,p,l,a} by: Anonymous --------------------------------------- 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: {c} {p} {l} {a} ----------------------------------- 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: {c,p} {c,l} {c,a} {p,l} {p,a} {l,a} ----------------------------------- 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: {c,p,l} {c,p,a} {c,l,a} {p,l,a} ----------------------------------- >>>The FINAL ANSWER: The number of all possible subsets (including the improper subset) of { c,p,l,a } = 15 (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: {c} {p} {l} {a} {c,p} {c,l} {c,a} {p,l} {p,a} {l,a} {c,p,l} {c,p,a} {c,l,a} {p,l,a} Improper Subset: { c,p,l,a } Thanks for writing. Staff www.solving-math-problems.com