the 2^n rule is applicable for finding the number of distinct genotypes of sperm or egg. The reasoning is that by independent assortment, a true heterozygote AaBbCc will have: (2 possible genotypes for A) x (2 possible genotypes for B) x (2 possible genotypes for C). You should exclude homozygous genes from the n count, as these only result in one possible genotype. If you want to look at the progeny, you would then multiply the number of possible genotypes or phenotypes of the sperm or eggs by looking at the minipunnet results of each gene.
If you were to cross AABBCC x aabbcc, you would only have ABC from parent1 and abc from parent2, leading to only one genotype/phenotype in the progeny, AaBbCc. The F2 generation would have 2^3=8 different genotypic gametes. Since each of the 3 heterozygous loci lead to 3 different genotypes representing 2 different phenotypes, the F2 generation would have 3^3 different genotypes comprising 2^3 different phenotypes.
Looking at the original question, AABbcc x aaBbCc. Parent 1 would yield only 2^1 genotypically distinct gametes using the 2^n rule: ABc and Abc, and Parent 2 would yield 2^2=4: aBC, abC, aBc, abc. The progeny then would have 1x3x2 different genotypes representing 1x2x2 phenotypes.
Given this, you should only use the 2^n rule for finding the number of genotypically distinct gametes from a parent, as this is what the rule is made for. Don't use it for finding phenotypes. If you play around with the numbers you can make any rule fit one example, like jpatel's. AaBb x AaBb gives 2^2=4 phenotypes. AA x Aa gives 2^1 =/= 1 phenotype, BUT aa x Aa gives 2^1=2 phenotypes. AND that doesn't take into account if Aa is a different phenotype as AA (in which case it would have to be the 3^n rule). Sooo, the proper conclusion is to not use the rule for situations it is not meant to be applied to.
