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How many two input AND gates and two input OR gates are required to realize Y = BD + CE + AB?

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How many two input AND gates and two input OR gates are required to realize Y = BD + CE + AB?

(a) 3, 2

(b) 4, 2

(c) 1, 1

(d) 2, 3

I had been asked this question during an online exam.

I need to ask this question from Logic Gates and Networks topic in section Boolean Algebra and Minimization Techniques and Logic Gates of Digital Circuits

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The correct option is (a) 3, 2

The best I can explain: There are three product terms. So, three AND gates of two inputs are required. As only two input OR gates are available, so two OR gates are required to get the logical sum of three product terms.

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