Please help me with this question sir,
C2A. Show that the complement of (a’ +b’)(a+c’)(b’+c’) is given by ab + a’c.
Write down the Boolean laws in each step of you simplification.
Show your working clearly.

b)Find out how many distinct 4-letter words can be formed from the word MESSAGE.

If you don’t know about Redundancy Theorem then you may use the solution below ab + a'c + bc.1 {since x.1 = x} ab + a'c + bc(a' + a) {since x’ + x = 1} ab + a'c + a'bc + abc {from Distributive Law} ab(1 + c) + a'c(1 + b) {taking ab common from the first and last terms and taking a’c common from the two middle terms} ab.1 + a'c.1 {since 1 + x = 1} ab + a'c {since 1.x = x}

The b part of the problem is not based on Boolean Algebra so please ask it in math section.

To understand this you must know about the AND operation. When you perform AND operation between any variable and 1 you will get the same variable, as the final result will only depend on the variable. For example, AND operation of A and 1 will be equal to A, when A is 0 we will have 0.1 which is 0 and when A is 1 we will have 1.1 which is 1. If you still have doubt in this you may watch Introduction to Boolean Algebra (Part 1)

i understand the redundancy theorem and also just to clarify
ab + a’c + bc(a’ + a) {since x’ + x = 1}
must the 1 translate to be the variable a or it can be like (b’+b) or (c’+c) etc