Consider the colour at the top. There are $4$ different choices for this: red, yellow, green and blue. For the right hand colour, this can be any of the other $3$ colours that have not yet been used. The bottom colour can then be either of the other $2$ colours. The left hand part then has to have the remaining colour.
This means there are $4 \times 3 \times 2 \times 1 = 24$ different ways to paint all four sections. This means there are $23$ other
ways to paint the tile.
This problem is taken from the UKMT Mathematical Challenges.