B4. In a room, there are five chairs on which five children are sitting. All the children stand up and sit on a different chair than the one they were sitting on. In how many ways can they now be seated?

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B4. Children A to E were sitting on the chairs in that order, which we denote as ABCDE. Suppose A sits on chair 2.

On the first chair:

B sits, then there are two possibilities for C, D, and E to sit on chairs 3, 4, and 5, namely BAECD and BADEC.

C sits, then there are three possibilities for B, D, and E to sit on chairs 3, 4, and 5, namely CABED, CAEBD, and CADEB.

D sits, then there are again three possibilities (compare C on chair 1).

E sits, then there are again three possibilities (compare C on chair 1).

So there are 11 possibilities where A sits on chair 2.

Similarly, there are 11 possibilities in the cases where A sits on chair 3, 4, or 5. In total, there are thus $4 \times 11=44$ possibilities where everyone sits on a different chair.

![](https://cdn.mathpix.com/cropped/2024_04_17_bf0efe45fea0def43891g-2.jpg?height=214&width=228&top_left_y=2252&top_left_x=1348)