The Gateway to Computer Science Excellence

First time here? Checkout the FAQ!

x

+1 vote

Simplify the following expression

AB’C + A’BC + A’B’C

Solution given is A’C + B’C can someone show me how?

AB’C + A’BC + A’B’C

Solution given is A’C + B’C can someone show me how?

+2 votes

:AB'C + A'BC +A'B'C

=AB'C+A'C(B+B')

=AB'C+A'C x 1

=C(AB'+A')

=C[(A+A')(B'+A')] [`Distributive Law XY+Z==(X+Z)(Y+Z)`

]

=C x 1 x (B'+A')

=A'C+B'C

+1 vote

Let $f(A,B,C)=A\cdot\overline{B}\cdot C+\overline{A}\cdot B\cdot C+\overline{A}\cdot\overline{B}\cdot C$

Take common from first and last term

$f(A,B,C)=\overline{B}\cdot C\cdot(A+\overline{A})+\overline{A}\cdot B\cdot C$

$f(A,B,C)=\overline{B}\cdot C\cdot1+\overline{A}\cdot B\cdot C$ ${\color{Red}{[X+\overline{X}=1]} }$

$f(A,B,C)=\overline{B}\cdot C+\overline{A}\cdot B\cdot C$

$f(A,B,C)=C(\overline{B}+\overline{A}\cdot B)$

$f(A,B,C)=C(\overline{B}+\overline{A})\cdot(\overline{B}+B)$ ${\color{Magenta} {[X+Y\cdot Z=(X+Y)\cdot(X+Z)]}}$

$f(A,B,C)=C(\overline{A}+\overline{B})\cdot 1$

$f(A,B,C)=\overline{A}\cdot C+\overline{B}\cdot C$

- All categories
- General Aptitude 1.8k
- Engineering Mathematics 7.3k
- Digital Logic 2.9k
- Programming & DS 4.9k
- Algorithms 4.3k
- Theory of Computation 6k
- Compiler Design 2k
- Databases 4.1k
- CO & Architecture 3.4k
- Computer Networks 4.1k
- Non GATE 1.4k
- Others 1.4k
- Admissions 596
- Exam Queries 577
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 18

49,532 questions

54,123 answers

187,320 comments

71,044 users