A-Level Probability: worked solution

Question

A pizzeria sells pizza by the slice, 8 slices making a whole pizza. (a) Three friends buy 4 Hawaiian slices, 1 pepperoni, 1 seafood, 1 cheese and 1 vegetarian slice. In how many ways can the slices be set in a circle with the 4 Hawaiian slices together? [1] (b) After each friend eats one Hawaiian slice, the remaining 5 slices are shared among the three so each gets at least one more; in how many ways? [3]

(a) The 4 Hawaiian slices must sit together, so glue them into a single block and treat that block as one object. That leaves 5 objects to arrange around the circle: the Hawaiian block, the pepperoni, the seafood, the cheese and the vegetarian slice. For n distinct objects in a circle the number of arrangements is (n − 1)!, not n!, because rotating a whole arrangement does not create a new one - fix one object and arrange the rest around it. (5 − 1)! = 4! = 4 × 3 × 2 × 1 = 24 The 4 Hawaiian slices are identical to one another, so shuffling them within the block produces nothing new; there is NO extra factor of 4!. Number of ways = 24 (b) Each of the 3 friends eats one Hawaiian slice, so 3 of the 4 Hawaiian slices are gone and 5 slices remain: 1 Hawaiian, 1 pepperoni, 1 seafood, 1 cheese and 1 vegetarian. All 5 remaining slices are different from each other, and the 3 friends are distinguishable people, so we are counting the ways to hand out 5 DISTINCT objects to 3 DISTINCT people with nobody left empty-handed. Without the restriction, each of the 5 slices can go to any of the 3 friends independently: 3⁵ = 243 Now remove the bad cases using the inclusion-exclusion principle. Subtract the cases where one named friend gets nothing: choose which friend in C(3,1) = 3 ways, and then all 5 slices go to the remaining 2 friends in 2⁵ = 32 ways: 3 × 32 = 96 Those subtractions have removed the cases where TWO friends get nothing twice over (once for each of the two), so add them back: choose the 2 empty friends in C(3,2) = 3 ways, and all 5 slices go to the one remaining friend in 1⁵ = 1 way: 3 × 1 = 3 Number of ways = 3⁵ − C(3,1)·2⁵ + C(3,2)·1⁵ = 243 − 96 + 3 = 150