build the func f=(ab+c`)` using min mux 2:1

Question

Anonymous · Accepted Answer

(ab +c`)` -> (ab)`(c) --> and(nand(a,b),c)

in 2:1 mux, nand(a,b) => inputs (.0(1), .1(B), .sel(A)) 
in 2:1 mux, and (a,b) => inputs (.0(0), .1(B), .sel(A))

so the final output of the circuit will be a combination of 
F = 2:1 mux(.0 (0) , .1([2:1 mux(.0(1), .1(B), .sel(A))], .sel(C))

Anonymous · Answer

(ab+c`)`=(ab)`c
let`s say that mux lets bit z choose from x or y so the output is zy+z`x
first mux a choose from 0,b to get ab then ab choose from 1,0 to get (ab)` then c choose from 0,(ab)` to get (ab)`c

Madhava · Answer

Both of these answers are confusing , can someone post a clear answer with picture and diagram neatly.

Madhava · Answer

This is the correct answer:

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F = 2:1 mux(.0(0), .1(2:1 mux(.0(1), .1(B`), .sel(A))], .sel(C))
It must be B bar not B

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