The minimum number of NAND gates required to implement A+AB'+AB'C ?Plz explain aslo...

There are two addition Terms and Three inputs

Inputs are a,b,c

for b to b’ conversion one NAND gate is required

For adding two numbers 3 NAND Gates are required

Therefore split the term as Y = a + ab’ + ab’c as

Z = b’ —- One NAND Gate

Z1 = ab’ is an and operation . this requires 3 NAND gates of 2 inputs

Z2 = ab’ + ab’c = Z1(1+c) = Z1 since 1 + c in digital is 1

theretofore Y becomes a + Z1 Totally 4 NAND gates of two input is the Minimal requirement

There are other different methods too for implementation of this function