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How many subsets of three elements each exist in a set of six elements?
Wiki User
∙ 12y ago
34 duh
* * * * *
Perhaps before adding "duh" the person providing the original answer should have been sure of his answer!
The correct answer is 6C3 = (6*5*4)/(3*2*1) = 20
Wiki User
∙ 12y ago
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How do you determine the list of all the subsets of sets?
If a set has N elements then it has 2N subsets. So you can see that a list of all subsets soon becomes a very big task. For reasonably small values of N, one way to generate all subsets is to list the binary numbers from 0 to 2N. Then, each of these represents a subset of the original set. If the nth digit is 0 then the nth element is not in the set and if the nth digit is 1 then the nth element is in the set. That will generate all the subsets.
What is subsets and proper subsets?
A set "A" is said to be a subset of "B" if all elements of set "A" are also elements of set "B".Set "A" is said to be a proper subset of set "B" if: * A is a subset of B, and * A is not identical to B In other words, set "B" would have at least one element that is not an element of set "A". Examples: {1, 2} is a subset of {1, 2}. It is not a proper subset. {1, 3} is a subset of {1, 2, 3}. It is also a proper subset.
Simplest form of matter that can exist under normal laboratory conditions?
tell whether each fraction is in simplest form. if not, write it in simplest form, tell whether each fraction is in simplest form. if not, write it in simplest form,
What is Distributive property of multiplication over addition?
The distributive property states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products together.
What three days of the week contain three vowels each?
3
How many subsets of three elements each can be made from a set of five elements?
(5 x 4 x 3)/(3 x 2) = 10
How many subsets does set A have if the set A has 3 elements?
In a subset each element of the original may or may not appear - a choice of 2 for each element; thus for 3 elements there are 2 × 2 × 2 = 2³ = 8 possible subsets.
How many subsets are in this set a b c d e f?
64. You can use Pascal's triangle to figure out how many subsets have no elements, one element, two elements and so on. For this particular one, you will have 6 subsets with one element, 15 with two, 20 with three, 15 with four, 6 with five and only one each of all six and none at all.
How many subsets of a set with 100 elements have more than one element?
It is 2^100 because each of 100 elements can either be in or out. By the way the answer is 2^100-101, because there is one subset with no elements at all (the empty set)!
How do you determine the list of all the subsets of sets?
If a set has N elements then it has 2N subsets. So you can see that a list of all subsets soon becomes a very big task. For reasonably small values of N, one way to generate all subsets is to list the binary numbers from 0 to 2N. Then, each of these represents a subset of the original set. If the nth digit is 0 then the nth element is not in the set and if the nth digit is 1 then the nth element is in the set. That will generate all the subsets.
How many subsets are there in a set?
The number of subsets of a given set, including the set itself and the empty set, is 2n. Easiest way to see why: to make a particular subset, for each element in the original set you either chhose it or you don't. There are thus two possibilities for each element, so 2n possibilities for all n elements.
Can you prove that nC0 nC1 nC2 nCn2n?
Consider a set, S, consisting of n elements.Then the number of subsets of S containing r objects is equal to the number of ways of choosing r elements out of n = nCr.ThereforenC0 + nC1 + nC2 + ... + nCn is the total number of subsets of S.Now consider each subset of S. Each element of S can either be in the subset or not in the subset. So for each element there are two choices. Therefore, n elements give 2^n possible subsets.Since these are two different approaches to the total number of subsets of S,nC0 + nC1 + nC2 + ... + nCn = 2^n.
How many subsets are in the set 12345?
Number of subsets with no members = 1Number of subsets with one member = 5.Number of subsets with 2 members = (5 x 4)/2 = 10.Number of subsets with 3 members = (5 x 4 x 3 /(3 x 2) = 10.Number of subsets with 4 members = (5 x 4 x 3 x 2)/(4 x 3 x 2) = 5.Number of subsets with 5 members = 1Total subsets = 1 + 5 + 10 + 10 + 5 + 1= 32.A set with n elements has 2n subsets. In this case n = 5 and 25 = 32.The proof in the case that n = 5 uses a basic counting technique which say that if you have five things to do, multiply together the number of ways to do each step to get the total number of ways all 5 steps can be completed.In this case you want to make a subset of {1,2,3,4,5} and the five steps consist of deciding for each of the 5 numbers whether or not to put it in the subset. At each step you have two choices: put it in or leave it out.
What are the 3 subset of a line define each subset?
A line has infinitely many subsets, not just three. Any collection of points on the line constitute a subset.
What is the purpose of the overlapping section in a Venn diagram?
The overlapping sections show elements that belong to each of the two (or maybe three) sets that overlap there.The overlapping sections show elements that belong to each of the two (or maybe three) sets that overlap there.The overlapping sections show elements that belong to each of the two (or maybe three) sets that overlap there.The overlapping sections show elements that belong to each of the two (or maybe three) sets that overlap there.
Prove that A contains N elements and the different subsets of A is equal to 2?
Assuming the question is: Prove that a set A which contains n elements has 2n different subsets.Proof by induction on n:Base case (n = 0): If A contains no elements then the only subset of A is the empty set. So A has 1 = 20 different subsets.Induction step (n > 0): We assume the induction hypothesis for all n smaller than some arbitrary number k (k > 0) and show that if the claim holds for sets containing k - 1 elements, then the claim also holds for a set containing k elements.Given a set A which contains k elements, let A = A' u {.} (where u denotes set union, and {.} is some arbitrary subset of A containing a single element no in A'). Then A' has k - 1 elements and it follows by the induction hypothesis that (1) A' has 2k-1 different subsets (which also are subsets of A). (2) For each of these subsets we can create a new set which is a subset of A, but not of A', by adding . to it, that is we obtain an additional 2k-1 subsets of A. (*)So by assuming the induction hypothesis (for all n < k) we have shown that a set A containing kelements has 2k-1 + 2k-1 = 2k different subsets. QED.(*): We see that the sets are clearly subsets of A, but have we covered all subsets of A? Yes. Assume we haven't and there is some subset S of A not covered by this method: if S contains ., then S \ {.} is a subset of A' and has been included in step (2); otherwise if . is not in S, then S is a subset of A' and has been included in step (1). So assuming there is a subset of A which is not described by this process leads to a contradiction.
Why does a set containing n element have 2n subsets?
I presume you meant 2^n (2 raised to the nth power), not 2*n (2 times n). That's answers.com's character set problem again (I trust, giving you the benefit of the doubt). The answer is that for each of the n elements, it is either in any particular subset or it isn't. Which elements are in and which are not in a subset defines the subset. So for example, if n is 3, say a, b, and c, there are 2 sub-collections of the set of all subsets: those containing a and those not containing a. In each of those sub-collections, there are 2 sub-collections based on whether they contain b, for a total of 4 (2*2) sub-collections. Finally, of each of these 4, there are 2 subsets: those containing c and those not containing c, for a total of 8 (2*2*2 or 2^3) subsets. Got it?
