What is the minimum number of gates required to implement the Boolean function (AB + C) if we have to use only 2-input NOR gates?

What is the minimum number of gates required to implement the Boolean function $(AB + C)$ if we have to use only 2-input NOR gates? 

(A) 2 

(B) 3 

(C) 4 

(D) 5 

(GATE 2009) 

Answer: (B) 3 

Explanation: 

$(A + BC) = (A + C)(B + C) = \big((A + C)' + (B + C)\big)'$ 

Therefore, two NOR gates is requiredto implement the given Boolean function. 

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