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)andfis not one–one} is
A
19
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B
19.00
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C
19.0
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Solution
C={f:A→B|2∈f(A)andfis 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. =3C2×3C1=9 Case-III : If f(x)=2for exactly one elements then total number of many-one function =3C1×3C1=9
Total =1+9+9=19