A-Level Probability: worked solution

Question

Aaron and Sandy do not get along and refuse to be in the same group. How many groupings of the twelve satisfy this restriction?

Step 1 - count all the groupings first, ignoring the restriction. The twelve people are split into three groups of four. Choose the members of the first group: C(12,4) ways. Choose the second group from the eight who are left: C(8,4) ways. The last four people fill the third group: C(4,4) = 1 way. Total = C(12,4) × C(8,4) × C(4,4) = 495 × 70 × 1 = 34650. (The same count written as a single expression is 12!/(4!4!4!) = 34650.) Step 2 - use complementary counting: count the groupings we must throw away. It is much easier to count the groupings where Aaron and Sandy ARE together and subtract them, than to count the good ones directly. Choose which of the three groups the pair sits in: 3 ways. That group still needs 2 more members, chosen from the other 10 people: C(10,2) = 45 ways. The remaining 8 people fill the other two groups of four: C(8,4) × C(4,4) = 70 × 1 = 70 ways. Groupings with Aaron and Sandy together = 3 × 45 × 70 = 9450. Step 3 - subtract. Groupings with Aaron and Sandy in different groups = 34650 - 9450 = 25200. Check by counting the good groupings directly. Put Aaron into a group: 3 choices. Put Sandy into one of the other two groups: 2 choices. The remaining 10 people now fill the 3, 3 and 4 empty places that are left: 10!/(3!3!4!) = 4200 ways. Total = 3 × 2 × 4200 = 25200, which agrees.