Mathematics 1228A/B Lecture : T1_w11_sol.pdf

• School: Western University
• Department: Mathematics
• Course: Mathematics 1228A/B
• Professor: Valuri

Document Summary

If u = {1, 2, 3, 4, 5, 6, 7}, a = {1, 3, 4} and b = {3, 5, 6}, nd ac b. Solution: ac contains all the elements of u which are not in a, so we have ac = {2, 5, 6, 7}. And ac b, the intersection of ac and b, contains only those elements which are contained in both ac and b at the same time. Ac b = {2, 5, 6, 7} {3, 5, 6} = {5, 6} Given that n(a b c) = 20, n(a b) = 10, n(ac b) = 15 and n(ac b c) = 40, nd n(ac b). Solution: we use the formula n(ac b) = n(ac) + n(b) n(ac b). Of course, for any sets s and t we know that n(s) = n(s t ) + n(s t c). Therefore we get n(ac b) = n(ac) + n(b) n(ac b) = 55 + 25 15 = 65.