Question

Let A = {x₁, x_2, …, 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 exists exactly three elements x in A such that f(x) = y₂, is equal to

(A)  14⋅ ⁷C₂

(B)  16⋅ ⁷C₃

(C)  12⋅ ⁷C₃

(D)  14⋅ ⁷C₃

Answer 1

Correct option  (d) 14⋅( ⁷C₃) 

Explanation :

We have A = {x₁, x₂, x₃, … x₇} and B = {y₁, y₂, y₃}. 3 elements in A having image y₂ can be chosen in ⁷C₃ ways. Now we are left with 4 elements in A which are to be associated with y₁ or y₃ i.e. each of 4 elements in A has 2 choices y₁ or y₃ i.e. in (2)⁴ ways. But there are 2 ways when one element of B will remain associated i.e. when all 4 are associated with y₁ or y₃. 

Therefore, required number of functions = ⁷C₃((2)⁴ - 2)

= 14⋅( ⁷C₃) .