Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

ABC' = 110.
A'BC = 011.
AB'C = 101.

Normally,
The three input truth table is,
A B C Y
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 1------>A'BC
1 0 0 1
1 0 1 1 ------>AB'C
1 1 0 1------->ABC'
1 1 1 1

Based on the expression, we consider the marked ones.
So, the answer is 3. Means 3 1's will come.

AB'C- 101 because B' (bar) so remaining are same 5+3+6=14 and 14 number in binary 1110 so number of 1s are 3.

I want a brief expression for the answer.

Bar take as 0.
So A Bbar C = 1 0 1 = 5.
:
:
:
5 + 3 + 6 = 14 = 1110 = 1 + 1 + 1 + 0 = 3.

Step 1:

Consider product inputs: A B C are 1 and A- B- C- as 0.

Step 2:

Resulting SOP (Sum of product) will look like.

101+011+110 = X.

5+3+6 = 14 i.e 1110.

Step 3:

So we have 3 1's in answer.

Answer: 3.

Can't understand please explain and how it is derived 5, 3, 6, 14 how?

Given function is true if the three minterms are true, so number of minterms is 3, so number of 1's is three.

5+3+6=14=1110

5+3+6=14=10110

ie there is three 1.