JEE Main 2020 — Functions Question with Solution

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 $$\frac{3!}{2!1!}$$ 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 $$\times$$ $$\frac{3!}{2!1!}$$ $$\times$$ 2! = 18

$$\therefore$$ Total number of functions = 1 + 18 = 19