Concept explainers
(A'B' + AB)(A'B + AB')(A + B)
Let's simplify the given expression using Boolean algebra:-
First, we can apply the distributive law to the first two terms:
(A'B' + AB)(A'B + AB')(A + B)
= (A'B'A' + A'BA)(A'B + AB') (A + B) // distribute A'B' over AB
= (A'A'B'B + A'BA'B + A'B'A' + ABA)(A + B') (A + B) // distribute A'B over AB'
= (0 + A'BA'B + A'B'A' + ABA)(A + B') (A + B) // A'A'B'B = 0 using A'A = 0 and B'B = 0
= (A'BA'B + A'B'A' + ABA)(A + B') (A + B) // identity law: 0 + X = X
Next, we can apply the distributive law again:
(A'BA'B + A'B'A' + ABA)(A + B') (A + B)
= (A'BA'B)(A + B') (A + B) + (A'B'A')(A + B') (A + B) + (ABA)(A + B') (A + B) // distribute over (A + B') and (A + B)
= (A'BA'BA + A'BA'BB' + ABA'A + ABA'B' + A'B'BA + A'B'BB' + ABA'A + ABA'B')(A + B) // distribute over (A + B)
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- ((A – B – C) U (B – A – C) U (C – A – B) – (AN B N C)arrow_forwardModule/Week 2 ASSIGNMENT (INPUT/OUTPUT)The number of permutations of a set of n items taken r at a time is given by the following formulan !⁄r !(n-r)!: where n! is the factorial of n, r! is the factorial of r, and (n-r)! is the factorial of the result of n-r. The factorial of a number n can be solved using the following formula: 〖n!=e〗^(-n) n^n√2πn. If there are 18 people in your class and you want to divide the class into programming teams of 3 members, you can compute the number of different teams that can be arranged using this formula (n !⁄r !(n-r)!). Write a C++ program that determines the number of potential team arrangements. You will need to use the double type for this computation. Use the Lab Template you set-up last week, proper formatting, and appropriate comments in your code. The output must be labeled clearly and formatted neatly. Submit C++ Programming Assignment 2 by 11:59 p.m. (ET) on Monday of Module/Week 2.arrow_forward2. Boolean Algebraarrow_forward
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