There are five blue chairs and four pink chairs in a class of nine students, comprising five boys and four girls. Before lunch, the five boys sat on the blue chairs and the four girls sat on the pink chairs. If after lunch, no girl sat on the same chair on which she was sitting before lunch, but all the girls sat on pink chairs, in how many ways can all the students be seated?
Question
There are five blue chairs and four pink chairs in a class of nine students, comprising five boys and four girls. Before lunch, the five boys sat on the blue chairs and the four girls sat on the pink chairs. If after lunch, no girl sat on the same chair on which she was sitting before lunch, but all the girls sat on pink chairs, in how many ways can all the students be seated?
Solution
The problem can be solved by using the concept of derangements, also known as the problem of counting permutations of the elements of a set that leave no element in its original place.
First, let's consider the girls. Before lunch, each girl was sitting on a unique pink chair. After lunch, no girl can sit on the same chair, so we need to find the number of derangements for 4 elements (the 4 girls).
The formula for derangements is !n = n! * (1/0! - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!).
So, for 4 elements, the number of derangements is !4 = 4! * (1 - 1/1! + 1/2! - 1/3! + 1/4!) = 9.
Now, let's consider the boys. After lunch, they can sit on any of the blue chairs, so there are 5! (factorial) ways to arrange them.
Therefore, the total number of ways all the students can be seated is the product of the number of arrangements for the girls and the boys, which is 9 * 5! = 9 * 120 = 1080.
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