LOTF Movie Choosing
In English Class, after studying a book called ``The Lord of the Flies" by William Golding, students are split into 6 groups and each group is required to make a film based on one of the 8 important scenes in the novel.
Each group can choose which scene they are creating a film of. However, to make sure there is a variety of scenes being created, there can only be a maximum of 2 groups creating the same scene.
Find the last 3 digits of the number of possible ways there are for each group to choose a scene.
Details and Assumptions
Each group has different students in it, so therefore each group is considered distinct.
Image credit: Wikipedia Vega
The answer is 400.
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1 solution
The correct answer is 600, not 400.
To solve the problem, we take cases, based on the number of scenes that get filmed twice.
Case 1 . No scene gets filmed twice.
The first group can choose one of 8 scenes, then the second group can choose one of 7 scenes (anything other than what the first group chose, because no scene is chosen twice), then the third group can choose one of 6 scenes, and so on, until the sixth group can choose one of 3 scenes. This gives 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 = 2 0 1 6 0 possible choices.
Case 2 . Exactly one scene gets filmed twice.
Let's say that two groups form a team if they film the the same scene. There are ( 2 6 ) = 1 5 ways to choose a team, and this team can choose one of 8 scenes.
This leaves four groups. Of these four groups, the first group can choose one of 7 scenes, and so on, until the fourth group can choose one of 4 scenes. This gives 1 5 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 = 1 0 0 8 0 0 possible choices.
Case 3 . Exactly two scenes get filmed twice.
There are ( 2 6 ) = 1 5 ways to choose a team (say team A), and this team can choose one of 8 scenes. There are then ( 2 4 ) = 6 ways to choose another team (say team B), and this team can choose one of 7 scenes. However, the order in which we choose the teams does not matter, so we must divide by 2 ! = 2 .
After the two teams are chosen, this leaves two groups. Of these two groups, the first group can choose one of 6 scenes, and the second group can choose one of 5 scenes. This gives 1 5 ⋅ 8 ⋅ 6 ⋅ 7 ⋅ 1 / 2 ⋅ 6 ⋅ 5 = 7 5 6 0 0 possible choices.
Case 4 . Exactly three scenes get filmed twice.
There are ( 2 6 ) = 1 5 ways to choose a team (say team A), and this team can choose one of 8 scenes. There are then ( 2 4 ) = 6 ways to choose a second team (say team B), and this team can choose one of 7 scenes. Then the two remaining groups must form the third team (say team C), and this team can choose one of 6 scenes. Again, the order in which we choose the teams does not matter, so we must divide by 3 ! = 6 . This gives 1 5 ⋅ 8 ⋅ 6 ⋅ 7 ⋅ 6 ⋅ 1 / 6 = 5 0 4 0 possible choices.
The total number of choices is then 2 0 1 6 0 + 1 0 0 8 0 0 + 7 5 6 0 0 + 5 0 4 0 = 2 0 1 6 0 0 .