34 duh
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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
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.
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.
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,
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.
The main reason is that many elements exist in several isotopic forms. These are atoms with the same number of protons and electrons but different numbers of neutrons. Since the atomic mass is essentially based on the number of protons and neutrons in each atom, the element will exist in several forms with different masses. The atomic mass is the average of the masses of these isotopes, weighted together according to their abundance.
(5 x 4 x 3)/(3 x 2) = 10
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.
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.
A set with ( n ) elements has ( 2^n ) subsets. This includes all possible combinations of the elements, including the empty set and the set itself. The reasoning behind this is that for each element, you can either include it in a subset or not, leading to ( 2 ) choices per element. Therefore, for ( n ) elements, the total number of subsets is ( 2^n ).
The subsets of the set {1, 2, 3, 4, 5, 6} include all possible combinations of its elements, including the empty set. There are a total of (2^6 = 64) subsets, which range from the empty set to the full set itself. Some examples of subsets are {1}, {2, 3}, {4, 5, 6}, and {1, 2, 3, 4, 5, 6}. Each subset can vary in size from 0 to 6 elements.
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)!
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.
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.
Well, honey, a set with 10 elements will have 2^10 or 1024 subsets in total. Now, if we want to count just the subsets with an odd number of elements, we need to consider that for each element, you have the option of including it or not. So, you're looking at 2^9 or 512 subsets with an odd number of elements. Hope that clears things up for ya, darling.
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.
A line has infinitely many subsets, not just three. Any collection of points on the line constitute a subset.
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.