JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 9)
Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the
number of elements in the set
C = {f : A $$ \to $$ B | 2 $$ \in $$ f(A) and f is not one-one} is ______.
C = {f : A $$ \to $$ B | 2 $$ \in $$ f(A) and f is not one-one} is ______.
Answer
19
Explanation
The desired functions will contain either one
element or two elements in its codomain of
which '2' always belongs to f(A).
Case 1 : When 2 is the image of all element of set A.
Number of ways this is possible = 1
Case 2 : When one image is 2 and other one image is one of {1, 3, 4}.
Number of ways we can choose one of {1, 3, 4} is = 3C1.
Now divide 3 elements {a, b, c} of set A into two parts.
We can do this $${{3!} \over {2!1!}}$$ ways.
Now map one part of set A into the element 2 of set B and map other part of set A into one of {1, 3, 4} of set B.
We can do that 2! ways.
So number of functions in this case
= 3C1 $$ \times $$ $${{3!} \over {2!1!}}$$ $$ \times $$ 2! = 18
$$ \therefore $$ Total number of functions = 1 + 18 = 19
Case 1 : When 2 is the image of all element of set A.
Number of ways this is possible = 1
Case 2 : When one image is 2 and other one image is one of {1, 3, 4}.
Number of ways we can choose one of {1, 3, 4} is = 3C1.
Now divide 3 elements {a, b, c} of set A into two parts.
We can do this $${{3!} \over {2!1!}}$$ ways.
Now map one part of set A into the element 2 of set B and map other part of set A into one of {1, 3, 4} of set B.
We can do that 2! ways.
So number of functions in this case
= 3C1 $$ \times $$ $${{3!} \over {2!1!}}$$ $$ \times $$ 2! = 18
$$ \therefore $$ Total number of functions = 1 + 18 = 19
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