A={a,b,c} B={1,2,3,4}. Then number of elements in set C= {f:A-->B 2€

1.	A={a,b,c} B={1,2,3,4}. Then number of elements in set C= {f:A-->B 2€f(A) and f is not one one}
Answer» C = {f : A → B 2 ∈ f(A) and f is not one-one} Case-I : If f(x) = 2 ∀ x∈ A, then number of function = 1 Case-II : If f(x) = 2 for exactly two elements then total number of many-one function = ³C₂³C₁ = 9 Case -III : If f(x) = 2 for exactly one element then total number of many-one function = ³C₁³C1 = 9 Total = 19

A={a,b,c} B={1,2,3,4}. Then number of elements in set C= {f:A-->B 2€f(A) and f is not one one}

Answer»

C = {f : A → B 2 ∈ f(A) and f is not one-one}

Case-I : If f(x) = 2 ∀ x∈ A, then number of function = 1

Case-II : If f(x) = 2 for exactly two elements then total number of many-one function

= ³C₂³C₁ = 9

Case -III : If f(x) = 2 for exactly one element then total number of many-one function

= ³C₁³C1 = 9

Total = 19

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A={a,b,c} B={1,2,3,4}. Then number of elements in set C= {f:A-->B 2€f(A) and f is not one one}

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