Solution : We have, P(n, r) = \(n!\over (n – r)!\) Putting r = n, \(\implies\) P(n, n) = \(n!\over 0!\) \(\implies\) n! = \(n!\over 0!\) [ \(\because\) P(n, n) = n! ] \(\implies\) 0! = \(n!\over n!\) = 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 \(\le\) n, then the total number of one-one functions from A to B is \(^nC_m \times m!\) where m! is m factorial. For example, Let set A have 3 elements and set B have 4 elements, then
What is the formula to find number of one one function ? Read More »
Solution : Let A be a set. Then, the relation \(I_A\) = {(a, a) : a \(\in\) A} on A is called the identity relation on A. In other words, then the relation \(I_A\) on A is called the identity relation if every element of A is related to itself only. Example : If A
What is the Identity Relation with Examples ? Read More »
Solution : Let A be a set. Then, A \(\times\) A \(\subseteq\) A \(\times\) 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
What are Universal Relation with Example ? Read More »
Solution : Let A be a set. Then, \(\phi\) \(\subseteq\) A \(\times\) 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
What is Void or Empty Relation with Example ? Read More »
Solution : Let A be a finite set containing n elements. Let 0 \(\le\) r \(\le\) 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 \(^nC_r\). Therefore, the number of subsets of A having
Prove that the total number of subsets of a finite set containing n elements is \(2^n\). Read More »
Solution : Let A be any set and \(\phi\) be the empty set. In order to show that \(\phi\) \(\subseteq\) A, we must show that every element of \(\phi\) is an element of A also. But, \(\phi\) contains no element. So, every element of \(\phi\) is in A. Hence, \(\phi\) 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 \(\in\) N and \(x^2\) = 9} is a singleton set equal to {3}. Note : The cardinal number of a singleton set is 1.
What is Singleton Set ? Read More »
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.
What is Cardinality of Set ? Read More »
Solution : The general solution of \(tan \theta\) = \(tan \alpha\) is given by \(\theta\) = \(n\pi + \alpha\), n \(\in\) Z. Proof : We have, \(tan \theta\) = \(tan \alpha\) \(\implies\) \(sin \theta\over cos \theta\) = \(sin \alpha\over cos \alpha\) \(\implies\) \(sin \theta cos \alpha\) – \(cos \theta sin \alpha\) = 0 \(\implies\) \(sin
What is the General Solution of \(tan \theta\) = \(tan \alpha\) ? Read More »
