Relations & Functions Question - None

Question

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}

Accepted Answer

Step 1: Find total functions f: A → B with 2 ∈ f(A).

Total functions from A to B = 4³ = 64.

Functions where 2 is NOT in the range = functions from A to {1,3,4} = 3³ = 27.

Functions where 2 IS in the range = 64 - 27 = 37.

Step 2: Find injective (one-one) functions f: A → B with 2 ∈ f(A).

Since |A| = 3 and |B| = 4, injective functions exist. Total injective functions = P(4,3) = 4!/(4-3)! = 4 × 3 × 2 = 24.

For injective functions with 2 ∈ f(A): 2 must be the image of exactly one element from A (say a₁), and the remaining 2 elements map injectively to 2 elements from {1,3,4}.

Choose which element of A maps to 2: 3 ways.

Choose 2 elements from {1,3,4} for remaining images: C(3,2) = 3 ways.

Arrange these 2 elements among remaining 2 elements of A: 2! = 2 ways.

Injective functions with 2 ∈ f(A) = 3 × 3 × 2 = 18.

Step 3: Apply complementary counting for non-injective functions.

Non-injective functions with 2 ∈ f(A) = (Functions with 2 ∈ f(A)) - (Injective functions with 2 ∈ f(A))

= 37 - 18 = 19.

∴ Answer: 19

Question:

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