We all know that a 2-input multiplexer selects one out of the available two inputs. Similarly, an N-input multiplexer selects one out of the available N inputs. Now, coming to the answer of the question,
The number of 2-input multiplexers needed to implement an N-input multiplexer is (N-1).
We will arrive at this conclusion by giving an example of an 8-input multiplexer implemented with the help of 2-input multiplexers. Each of the 2-input muxes reduces the number of signals by 1, thereby requiring total of "7" 2-input muxes to implement an 8-input mux.
Let us consider a circuit with 8 inputs and variable outputs. Firstly, if it has 8 outputs, there is no mux in-between.
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| Figure 1: Circuit with 8 output lines to represent 8 input lines |
Now, if we add a 2-input mux between any of the 2 lines, then, we are reducing the number of output lines by 1 as shown in figure 2.
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| Figure 2: Circuit with 7 output lines and 8 input lines |
Simlarly, adding another 2-input mux will leave the number of output lines to be 6. Figure 3 shows two of the all possible configurations of such circuits.
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| Figure 3: Circuits with 6 output lines and 8 input lines |
Similarly, continuing on similar lines, we will need "7" 2-input muxes to converge to a single output. Thus, we can say that "7" 2-input muxes make an 8-input mux.
Similarly, going by this, we will need 13 2-input muxes for a 14:1 mux, 31 for a 32:1 mux, and so on. In other words, (N-1) 2:1 muxes will make up an N-input mux.