Puzzles

One method for putting the five balls into the four boxes is: pick one of the balls (there are 5 ways to do this), then put the other four balls into boxes with one in each box (there are 4! ways to do this), then put the picked ball into a box (there are 4 ways to do this). This method gives exactly two ways of obtaining each possible arrangement, so the number of ways is \(5\times4!\times4\div2\) or 240.

The two tokens must be in two non-adjacent columns. There are ½×21×20 ways of picking two different columns. 20 of these ways will give two adjacent columns, so there are ½×21×20–20=190 ways to pick the columns.

Once the columns are picked there are four choices for the rows to place the tokens in (up and up, up and down, down and up, down and down). 4×190=760.

Either both pieces must be on the diagonal, or one pieces is in the lower right half and the other is in the reflected position in the upper right half.

There are \(\left(\begin{array}{c}29\\2\end{array}\right)=406\) ways to pick two squares on the diagonal. There are 406 squares below the diagonal.

Therefore there are 406+406 = 812 ways to arrange the pieces.

First, consider placing 5 tiles in a 1×9 grid. There is only one way to do this:

To get the number of ways of placing 5 tiles in a 1×10 grid, imagine adding an extra blank square to either the start or end of the grid or between two of the counters. There are 6 places this tile could be inserted leading to 6 arrangements of 5 tiles in a 1×10 grid.

For 5 tiles in a 2×10 grid, you can first pick the columns the tiles go in (as a tile being in a column means nothing can be placed the columns either side, the number of ways to pick columns is the same and the number of wats to arrange 5 tokens in a 1×10 grid). For each of these column choices, there are two locations for each tile (top or bottom). This leads to a total number of arrangements of 6×2⁵=192.