As we know that mux is a fully functionally complete circuit(It can implement any Boolean function).

we know that if I have 4 *1 mux than I can implement any function with 2 variables.

and if I have 8*1 mux then I can implement any function with 3 variables but how???

Actually, in 8 *1 mux, I have 3 select lines and 8 inputs and these 8 inputs are actually minterms(or combinations possible with 3 variables). So, I can assign a value to each of 8 Combinations based on the select lines.

So, in the same way, I can implement an "n" variable function with the help of 2^n *1 MUX.

But now the actual question come that is: can I implement a 3 variable function with 4*1 MUX or Can I implement a "n+1" variable function with 2^n * 1 MUX????

So, here the answer is yes, I can implement 3 variable functions with 4*1 MUX OR I can implement "n+1" or more than this with 2^n *1 MUX. But how, so I will show u with example

I have explained in above page in a very easy way. In the same way, I can implement more than "n" variable function with 2^n * 1 MUX.

So what I need to do is, for implementing 3 variable functions with 2^2 * 1 MUX, I need to give(3-1) select lines and one output line.

In the same go, u know that they have given 1 multiplexor and one, not gate, for implementing a function of "n" variable and they are asking about what would be a minimum size of a multiplexor, so out of n variables, "n-1" will be select lines and one output line.

So minimum size of multiplexor = 2^(n-1) * 1

So if anyone has any kind of doubt then feel free to ask me...