Prove that value of zero factorial is 1.
Solution : We have, P(n, r) = n!(n–r)! Putting r = n, ⟹ P(n, n) = n!0! ⟹ n! = n!0! [ ∵ P(n, n) = n! ] ⟹ 0! = n!n! = 1 Hence, Proved.
Prove that value of zero factorial is 1. Read More »
Solution : We have, P(n, r) = n!(n–r)! Putting r = n, ⟹ P(n, n) = n!0! ⟹ n! = n!0! [ ∵ P(n, n) = n! ] ⟹ 0! = n!n! = 1 Hence, Proved.
Prove that value of zero factorial is 1. Read More »
Solution : If A and B are two sets having m and n elements respectively such that m ≤ n, then the total number of one-one functions from A to B is nCm×m! where m! is m factorial. For example, Let set A have 3 elements and set B have 4 elements, then
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Solution : Let A be a set. Then, the relation IA = {(a, a) : a ∈ A} on A is called the identity relation on A. In other words, then the relation IA on A is called the identity relation if every element of A is related to itself only. Example : If A
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Solution : Let A be a set. Then, A × A ⊆ A × A and so it is a relation on A. This relation is called the universal relation on A. In other words, a relation R on a set is called universal relation, if each element of A is related to every element
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Solution : Let A be a set. Then, ϕ ⊆ A × A and so it is a relation on A. This relation is called the void or empty relation on set A. In other words, a relation R on a set A is called void or empty relation, if no element of A is
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Solution : Let A be a finite set containing n elements. Let 0 ≤ r ≤ n. Consider those subsets of A that have r elements each. We know that the number of ways in which r elements can be chosen out of n elements is nCr. Therefore, the number of subsets of A having
Prove that the total number of subsets of a finite set containing n elements is 2n. Read More »
Solution : Let A be any set and ϕ be the empty set. In order to show that ϕ ⊆ A, we must show that every element of ϕ is an element of A also. But, ϕ contains no element. So, every element of ϕ is in A. Hence, ϕ is the subset of A.
Prove that Empty Set is a Subset of Every Set. Read More »
Solution : A set consisting of a single element is called a singleton set. Example : The set {5} is a singleton set. Example : The set {x : x ∈ N and x2 = 9} is a singleton set equal to {3}. Note : The cardinal number of a singleton set is 1.
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Solution : The cardinality of a set is the number of elements in a set. For example : Let A be a set : A = {1, 2, 4, 6} Set A contains 4 elements. Therefore, Cardinality of set is 4.
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Solution : The general solution of tanθ = tanα is given by θ = nπ+α, n ∈ Z. Proof : We have, tanθ = tanα ⟹ sinθcosθ = sinαcosα ⟹ sinθcosα – cosθsinα = 0 ⟹ \(sin
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