Consider a gene with four alleles A1, A2, A3, and A4. How many distinct heterozygous genotypes are possible?

This can be solved either by using a Punnet Square or you can use the formula (n (n+1)) / 2, where n is the number of alleles. However, using either method will give you the number of possible genotypes regardless if it's homozygous or heterozygous. If we use the second formula, that will give you 10 possible genotypes. Since you only want to know the number of heterozygous genotypes, you should subtract the number of possible homozygous genotypes. In this case, there are four. That is (A1, A1), (A2, A2), (A3, A3), and (A4, A4). Knowing this, you'd be able to figure out that there are six heterozygous genotypes.

(A1, A2), (A2, A3), (A3, A4), (A1, A4), (A1, A3), (A2, A4)