• a = houses only
• b = houses and cars, but no boats
• c = cars only
• d = houses and boats, but no cars
• e = all three
• f = cars and boats, but no houses
• g = boats only
• h = none of the three items mentioned

(a)  You are given a universal set of 900 workers and three its basic subsets

         - H (owned houses) of 600 workers;
         - C (owned cars)   of 500 workers;
         - B (owned boats)  of 345 workers.

     You also are given the info about in-pair intersections of these three subsets

         - CH (owned cars   and houses)  of 300 workers;
         - HB (owned houses and boats)   of 250 workers;
         - CB (owned cars   and boats)   of 270 workers.

     You also are given the info about the triple intersections of these three subsets

         - CHB (owned all three items)  of 200 workers.


     Use the Inclusion-Exlusion principle.  The Inclusion-Exclusion formula is

         n(H U C U B) = n(H) + n(C) + n(B) - n(HC) - n(HB) - n(CB) + n (CHB) = 

                      = 600 + 500 + 345 - 300 - 250 - 270 + 200 = 825.

     Thus 825 workers own at least one of three items.

     Hence, the rest  1000-825 = 175  workers don't have any of items.    ANSWER



(b)  To find the number of workers who own only two of the items, subtract triple intersection from 
     the corresponding in-pair intersection.

     You will get:

         n(CH_only) = n(CH) - n(CHB) = 300 - 200 = 100  (owned cars   and houses only);

         n(HB_only) = n(HB) - n(CHB) = 250 - 200 =  50  (owned houses and boats only);

         n(CB_only) = n(CB) - n(CHB) = 270 - 200 =  70  (owned cars   and boats only).

     These sets CH_only, HB_only and CB_only are disjoint (i.e. have empty intersection;  THEREFORE

     the number of workers owned only two of the items is the sum  100 + 50 + 70 = 220.    ANSWER