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Let \(\mathrm{A}=\{1,2,3,4\}\) and \(\mathrm{B}=\{1,4,9,16\}.\) Then the number of many-one functions \(f: A \rightarrow B\) such that \(1 \in f(\mathrm{~A})\) is equal to :

(1) 127

(2) 151

(3) 163

(4) 139

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Best answer

Correct option is (2) 151 

\(Total =4^{4}\)

One-one =4!

Many-one = 256 - 24 = 232

Many-one which \(1 \notin \mathrm{f}(\mathrm{A})\)

\(=3 \cdot 3 \cdot 3.3=81\)

232 - 81 = 151

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