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If only two bit vectors are allowed to use in the VHDL code, then how many number of MUX will be required to implement 4 to 1 MUX?

If only two bit vectors are allowed to use in the VHDL code, then how many number of MUX will be required to implement 4 to 1 MUX? Correct Answer 3

Since we have inputs with two bits only, so we can use 2 to 1 MUX to implement the required design. So, to design 4 to 1 MUX, we need 3 2 to 1 MUX and hence we can get the desired circuit by using 3 multiplexers.

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