# Let A = {x1, x2, …, x7} and B = {y1, y2, y3} be two sets containing seven and three distinct elements respectively.

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Let A = {x1, x2, …, 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

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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) .

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