Let A=(a,b,c) and B=(1,2,3,4). Then the number of elements in the set C={f: A→B | 2∈f(A) and f is not one-one} is
Let A=(a,b,c) and B=(1,2,3,4). Then the number of elements in the set C={f: A→B | 2∈f(A) and f is not one-one} is
19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? * (3!/(2!1!)) * 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19
Similar Questions for you
? P? = 6!/2! = 360
Kindly consider the following Image
(on simplification)
Use [x + n] = n + [x], where
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...

Maths Relations and Functions 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering
