Let A & B be two sets containing four & two elements. Then, what will be the number of subsets of set A×B each having at least three elements?
Let Set A = {1,2,3,4} & Set B = { a,b}=> A x B contains 4*2 = 8 elements ( or ordered pairs),Like A x B = { (1,a),(1,b),(2,a),(2,b),(3,a),(3,b),(4,a)(4,b) }So, A x B contains 8 ordered pairs.Now we need to find the subsets of A x B , in which at least 3 elements ie 3 ordered pairs should be there. That means it can have 8 ordered pairs or 7 ordered pairs or 6pairs or 5pairs or 4pairs or 3 pairs. It can not go below 3 pairs as question is we need subsets with at least 3 pairs.So we start making the subsets with 8 pairs.●With 8 ordered pairs…. we have just 1 subset , which is {(1,a),(1,b),(2,a,),(2,b),(3,a),(3,b),(4,a),(4,b)} (as such , every set is its own subset)●similarly with 7 ordered pairs : the number of ways of choosing 7 pairs from the set of 8 pairs will be= 8C7 = 8!/(1! * 7!) = 8 subsets● Now similarly with 6 pairs : we get 8C6= 8!/(2! * 6!) = 28 subsets● With 5 pairs : we get 8C5 = 8!/(3!*5!) = 56 subsets● With 4 pairs : we get 8C4 = 8!/(4!*4!) = 70 subsets●Now with 3 pairs: we get 8C3 = 8!/(5!*3!) = 56 subsetsNow, by adding all the above subsets , we get..1 + 8 + 28 + 56 + 70 + 56 = 219 subsets . . . . Ans
Two finite sets have m and n elements each. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. What are the values of m and n?
Number of Subsets = [math]2^n[/math]Difference in number of Subsets should be of the form [math]2^x(2^y-1)[/math][math]56 = 8 \times 7 = 2^3 \times (2^3–1)[/math]n = y = 3m = x+y = 6Detailed ExplanationStep 1 : Property of SubsetsCardinality : A set of [math]n[/math] elements will have have [math]2^n[/math] subsets.This means Set A of m elements have [math]2^m[/math] subsets.Again Set B of n elements have [math]2^n[/math] subsets.Step 2 : Set A have 56 more subsets than Set B.It is means [math]2^m - 2^n = 56[/math]Step 3 : Factorizing 56It is clear than m is greater n.Let m = n + k,[math]2^m-2^n = 56[/math][math]2^n (2^k-1) = 56[/math][math]2^k-1[/math] is always an odd number. So we need to find the odd factors of 56.[math]56 = 2 \times 2 \times 2 \times 7 [/math] → 7And fortunately, 56/7 = 8 with is a power of 2.If not the question would have been wrong !Post Check[math]56 = 7 \times 8 = (2^3–1) 2^3[/math]n = 3 and k = 3 → m = n + k = 6[math] 2^6 = 64[/math][math] 2^3 = 8 [/math]64–8 = 56
If A and B are two sets containing m and n distinct elements respectively, how many different relations can be defined A and B?
A relation is a set of ordered pairs (a,b) where a ∈ A and b ∈ B, the number of such ordered pairs will be m*n , each subset of those pairs will be a relation. The number of subsets will be 2^mn
Sets A and B have 3 and 6 elements each. What is the minimum number of elements in (A U B)?
The cardinality of the union between sets A and B could be anywhere between 6 and 9 elements. Remember that a union between A and B implies the combination of elements in sets A and B. If we know that one set is a subset of the other, the cardinality of their union will be the greater cardinality of the two sets.Example 1 (Your Question):Sets A and B are very similar because set A is actually a subset of set B. In other words, all members of set A are present in set B.[math]A = \{ a, b, c \}[/math][math]B = \{ a, b, c, d, e, f \}[/math][math]C = A \cup B[/math][math] = \{ a, b, c \} \cup \{ a, b, c, d, e, f \}[/math][math] = \{ a, b, c, d, e, f \}[/math]Here, because A was a subset of B, the cardinality of their union was 6.Example 2:Sets A and B share some similiar elements but not all of them. In other words, A is not a subset of B but has some of the same elements.[math]A = \{ a, b, c \}[/math][math]B = \{ b, c, 1, 2, 3, 4 \}[/math][math]C = A \cup B[/math][math] = \{ a, b, c \} \cup \{ b, c, 1, 2, 3, 4 \}[/math][math] = \{ a, b, c, 1, 2, 3, 4 \}[/math]Here, because two elements of A were also members of B, only 1 new element was introduced to B. The cardinality of their union was 7.Example 3:Sets A and B are in no way, shape, or form related. In other words, both sets are completely unique.[math]A = \{ a, b, c \}[/math][math]B = \{ 1, 2, 3, 4, 5, 6 \}[/math][math]C = A \cup B[/math][math] = \{ a, b, c \} \cup \{ 1, 2, 3, 4, 5, 6 \}[/math][math] = \{ a, b, c, 1, 2, 3, 4, 5, 6 \}[/math]Here, because A and B were unique from each other, the cardinality of their union was 9.
If A and B are disjointed sets, how can you find n (A union B)?
Two sets are said to be disjoint if they have no element in common. Equivalently, disjoint sets are sets whose intersection is the empty set. For example, {1, 2, 3} and {4, 5, 6} are disjoint sets,( as they have no element in common) while {1, 2, 3} and {3, 4, 5} are not.(as the element 3 is in common).The union of two sets A and B is the set of elements which are in A, in B, or in both A and B. In figures;First consider the case where the sets A and B are disjoint.In that case,The number of elements in the union (A∪B) is simply the sum of the number of elements in A and the number of elements in B: |A ∪ B| = |A| + |B|. [ |A|→no of elements in A and other notations mean similar].But if A and B overlap, then the latter formula does not hold because, we are counting the elements in the intersection (A ∩ B) twice. Compensating for that leads to the given formula: |A ��� B| = |A| + |B| − |A ∩ B|.[ Note : n(A U B) is also denoted as |A U B| ]In above figure of Disjoint sets;Elements in (AUB)=elements in (A)+ elements in (B).In above example, union of disjoint sets is;Element set in A + Element set in B={1,2,3,5,7,9}.
If A and B are sets,where [math]n(AUB)=36, n(A∩B)=16, n(A-B)=15[/math] then find n(B)?
The question has already been answered correctly. Let me try a different take on this.Being sure of whats what can help you figure this solution easily...:)n(AUB) means the the number of elements in both A & B. And that includes the ones in common too.n(A∩B) means the ones that in common.n(A-B) refers to the elements that belong to A alone. That means the elements in A excluding the common elements.n(B) is the number of elements that belong to B and that includes the common ones too.Now coming to your question.The total number of elements that can fit in both the sets is 36.While the number of common elements is 16.And also given that n(A-B)=15.Now according to the definitions above....n(A) should include the ones that belong exclusively to A alone and also the ones in common.So,n(A)=n(A-B)+n(A∩B) =15+16 =31Therefore, n(A) is 31.Now when n(AUB) is 36 and n(A) is 31, it is clear that the elements exclusive to B alone is 5, i.e.36-31. Now n(B)=n(A∩B)+n(B-A)Therefore, n(B) is 16+5 = 21.n(B)=21.
If A = {1, 2} and B = {3, 4}, then how many subsets will A × B have?
GIVEN: A = {1,2}, B = { 3,4 }TO FIND: The subsets of A x B (A cross B)A x B = { (1,3), (1,4), (2,3), (2,4) }So, A x B has 4 elements or 4 ordered pairs.If we represent these elements or ordered pairs as p, q, r, sS = { p,q,r,s }Its subsets are{ p}, { q } , {r} , {s} ,{ p,q} , { p,r}, {p, s} , { q,r} , { q,s} , { r,s }{ p, q, r} , { p, r, s} , { q,r,s} , { q, s, p}{ p, q, r, s} &Null setthe formula: no of subsets of A x B = 2^n, where n is the number of elements or ordered pairs belonging to A x BSo, the above A x B has 16 subsets
How do you find n (AnB) with given n (A) and n (B) and n (AuB)?
There is a specific formula that will suit your need or satisfy itn(AuB) represents the total no. of elements in A and B, by counting the common elements in A and B only oncevoila! That derives our formulawe can say that if we count both the sets together our result will be equal to n(A) + n(B) which is equal to p(A) + p(B) + 2[n(AnB)], where p(X) denotes elements native to Only XTherefore, in order to get n(AuB), we subtract n(AnB) from the sum of n(A) + n(B)Let me express this as an equationS =p(A) + p(B) + 2[n(AnB)]S =n(A) + n(B)p(A) + p(B) + [n(AnB)] = n(AuB)S-n(AnB)=n(AuB)Hence we can now say that,n(A) + n(B) - n(AnB)= n(AuB)hope it helps
If set A = {1,2,3,4} & set B = {4,5,6}, then what is set A-set B?
Given set A = {1,2,3,4} & set B = {4,5,6}, then what is set A-set B = {1,2,3,4}-{4,5,6} ={1,2,3}Algebra
If n(A) = 7, n (A ∪ B) = 11, and n(B) = 5, then what is n (A ∩ B)?
Generally for 2 different sets A and B UNION operation(A u b)includes the common elements A and B once, while INTERSECTION (A n B) deals with distinct elements of A and B.Hence sum of no. of elements in A(n(A)) and no. of elements in B (n(B)) can be given asn(A)+n(B)=n(AuB)+n(AnB)Thus to find n(AnB)n(AnB)=n(A)+n(B)-n(AuB)=7+5-11=12Hence n(AnB)=1