Set, Relation and FunctionHard
Question
Let A = {x1, x2,......, x7} and B = {y1, y2, y3} be two sets containing seven and three distinct elements respectively.Then the total number of functions f : A → B that are onto, if there exist exactly three elements x in A such that f(x) = y2, is equal to :
Options
A.14.7C3
B.16.7C3
C.14.7C2
D.12.7C2
Solution
Number of onto function such that exactly three elements in x → A such that f(x) = 1/2 is equal to
= 7C3.{24 − 2}
= 14.7C3
= 7C3.{24 − 2}
= 14.7C3
Create a free account to view solution
View Solution FreeMore Set, Relation and Function Questions
Let f: R → R be a function such that f(2 − x) = f(2 + x) and f(4 − x) = f(4 + x), for all x ∈ R ...The number of relations, defined on the set $\{ a,b,c,d\}$, which are both reflexive and symmetric, is equal to:...If f is a real valued function defined on [0,1] such that f(0) = f(1) = 0 and |f(x1) - f(x2)| 1 - x2| for all distinct x...If for three disjoint sets A, B, C; n(A) = 10, n(B) = 6 and n(C) = 5, then n(A ∪ B ∪ C) is equal to -...A function f from the set of natural numbers to integers defined by f(n) = is...