Probability of having an all-heterogeneous genotype 
Find the probability that the genotype of an offspring will be $\mathrm{AaBbCcDdEeFfGgHhIiJj}$ from the cross $\mathrm{AaBbCcDdEeFfGgHhIiJj \times AaBbCcDdEeFfGgHhIiJj}$.

Making a Punnett square for this problem will be too tedious. However, I noticed that this problem can be treated as a combinatorics problem. For the simplest case, the possible outcomes of the cross $\mathrm{Aa \times Aa} = \{\mathrm{AA, Aa, Aa, aa}\}$ where two of them are desired. As there are 10 traits, we have $2^{10}$ of this genotype from the cross. Now, considering the number of genotypes that are not of this form, there are $2^1 + 2^2 + \cdots + 2^{10} = 2^{11} - 2$. Getting the number of all possible genotypes, we have
\begin{align*}2^{11} + 2^{10}-2 &= 2^{10}(2 + 1) - 2 \\ &= 3(2^{10}) - 2\end{align*} Therefore, the probability of getting the genotype is $$\frac{2^{10}}{3(2^{10}) - 2} = \frac{2^9}{3(2^9) - 1}$$
Can someone check if I overcounted or undercounted something? If not, is there a better way to solve these types of problems?
 A: There are four possible outcomes for each of the ten traits.  For instance, for the first trait, they are $AA, Aa, aA, aa$, where we first list the allele from the father, then the allele from the mother.  Since we are not told the traits are linked in some way, it is reasonable to assume independence. Thus, there are $4^{10}$ possible genotypes.
As you observed, there are two possible heterozygous outcomes for each trait.  For instance, for the first trait, they are $Aa, aA$.  Again, assuming independence, there are $2^{10}$ possible favorable outcomes.
Thus, the probability that the genotype of an offspring of two parents who are heterozygous for each of the ten traits is heterozygous for each of the ten traits is
$$\frac{2^{10}}{4^{10}} = \left(\frac{2}{4}\right)^{10} = \left(\frac{1}{2}\right)^{10} = \frac{1}{2^{10}} = \frac{1}{1024}$$
Where did you make your mistake?
The number of genotypes that are not heterozygous for all traits is $4^{10}−2^{10}$. It looks like you are saying that there are $2^k$ genotypes that different from those that are heterozygous for all traits in $k$ traits. However, that is not true. There are $$10⋅2^k⋅2^{10 - k} = \binom{10}{k} \cdot 2^{10}$$
ways for exactly one of the traits to be homozygous since there are $\binom{10}{k}$ ways to choose exactly $k$ of the traits to be homozygous, $2$ ways to obtain a homozygous version of that trait, and $2$ possible ways for each of the remaining $10 - k$ traits to be heterozygous.  Notice that
$$\sum_{k = 1}^{10} \binom{10}{k} 2^{10} = 4^{10} - 2^{10}$$
