Given three sets A B and C Suppose we know that the union of

Given three sets A, B, and C. Suppose we know that the union of the three sets has cardinality 220. Further, |A| = 120, |B| = 160, |C| = 150. Also, |A Intersection B| = 70, |A Intersection B| =80, and |B Intersection C| = 90. Find the following cardinalities |B - C|. |A Intersection (B Union C)|. Find a bag X that satisfies both of the following two simultaneous bag equations, [a, a, b, c, c, d, d, d] Intersection X = [a, c, c, d, d]. [a, a, b, c, c, d, d, d] Union X = [a, a, b, c, c, c, d, d, d].

Solution

1. a) |A|=120, |B|=160, |C|=150, |AB| = 70, |AC|= 80, |BC|= 90

|B-C| means number of elements of B which are not elements of C

|B-C|=|B|-|BC| = 160-90=70

b) |A(BC)| = |AUB| + |AC|

|AUB| = |A| +|B| -|AB| = 120+160-70 = 210

|A(BC)| = |AUB| + |AC| = 210+80 = 290

2. A= {a,a,b,c,c,d,d,d}

AX={a,c,c,d,d} and AUX= {a,a,b,c,c,d,d,d}

(AUX)-A = { }

((AUX)-A) U ( AX))= { } U {a,c,c,d,d}

X=((AUX)-A) U ( AX)) = {a,c,c,d,d}