Half and Half
Two of the four small triangles are to be painted black. In how many ways can this be done?
Problem
The large equilateral triangle is fixed in position. Two of the small equilateral triangles are to be painted black and the other two are to be painted white. In how many ways can this be done?
If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.
Student Solutions
Answer: 6
Here are two possible ways of thinking about this problem, although there are many other possibilities.
Choosing the colour of the top triangle
If the top triangle is painted black, there are three choices for the other black triangle, then the other two are painted white.
If on the other hand the top triangle is painted white, there are three choices for the other white triangle, with the other two painted black. This gives a total of
$6$ possibilities.
Choosing which triangles are black
If the top triangle is painted black, there are three choices for the other black triangle.
If not and the left hand triangle is painted black, the other black triangle could be the centre or right-hand triangles, so $2$ choices
If the top and left are both white, the other two triangles are both black.
This means there are $3+2+1=6$ possibilities.
This problem is looking at the number of ways to choose two of the four triangles to paint black. This can be written as $^4C_2$ or $4 \choose 2$, which are called binomial coefficiants. If you want to find out more about these, then you can look at this article.