If A{2, 3, 4, 5), B = {1,2,3} and C = {5,6}, then verify n(AUBUC) n(AN

1.	If A{2, 3, 4, 5), B = {1,2,3} and C = {5,6}, then verify n(AUBUC) n(ANB)+n(BNC)+n(CNA) = n(A) + n(B) + n(C)+n(An BNC)
Answer» Step-by-step EXPLANATION: Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6} For the LHS Union of two sets will have the ELEMENTS of both sets. So, B∪C={2,3,4,5,6,8} A−(B∪C) will have elements of A which are not in (B∪C) So, A−(B∪C)={1} (1) For the RHS A−B will have elements of A which are not in B. So, A−B={1,3,5} A−C will have elements of A which are not in C. So, A−C={1,2} Intersection of two sets has the common elements of both the sets. (A−B)∩(A−C)={1} (2) From (1) and (2), we have A−(B∪C)=(A−B)∩(A−C)

If A{2, 3, 4, 5), B = {1,2,3} and C = {5,6}, then verify n(AUBUC) n(ANB)+n(BNC)+n(CNA) = n(A) + n(B) + n(C)+n(An BNC)

Answer»

Step-by-step EXPLANATION:

Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}

For the LHS

Union of two sets will have the ELEMENTS of both sets.

So, B∪C={2,3,4,5,6,8}

A−(B∪C) will have elements of A which are not in (B∪C)

So, A−(B∪C)={1} (1)

For the RHS

A−B will have elements of A which are not in B.

So, A−B={1,3,5}

A−C will have elements of A which are not in C.

So, A−C={1,2}

Intersection of two sets has the common elements of both the sets.

(A−B)∩(A−C)={1} (2)

From (1) and (2), we have

A−(B∪C)=(A−B)∩(A−C)