Genetics and probability (“other than” case) (Moved from Bio SE due to the mathematical nature of the problem.)

If a man and woman, both carriers of a autosomal recessive
  disorder (i.e. having genotype $Aa$), produce three children, what is the probability of one or
  more children having the disorder?

The answer key suggests finding the probability of the children all being normal and then subtracting that from 1:
$$Aa \times Aa$$
$$P_{AA\:or\:Aa}=\frac{3}{4}$$
$$\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}\tag{3 normal}$$
$$1-\frac{27}{64}=\frac{37}{64}\tag{≥1 abnormal}$$
I understand the logic behind that: If they aren’t all normal, then at least one of them has to be abnormal.
However, what if I didn’t want to use the $1-x$ method? What if I wanted to do it the hard way?
I tried breaking it into three cases: 1 abnormal, 2 abnormal, and all abnormal.
$$P_{aa}=\frac{1}{4}$$
$$\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}\tag{1 abnormal, 2 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}\tag{2 abnormal, 1 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\tag{3 abnormal, 0 normal}$$
$$\sum P=\frac{13}{64}$$
which obviously isn’t the correct answer. What’s wrong with the way I’m doing it?

$$P_{aa}=\frac{1}{4}$$
$$\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}\tag{1 abnormal, 2 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}\tag{2 abnormal, 1 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\tag{3 abnormal, 0 normal}$$
The error in your argument lies in the first and second statements. Suppose the three children are $A,B,C$.
In the first case, where 1 is abnormal and 2 are normal, we can have,
Abnormal - $A$, Normal - $B,C$
Abnormal - $B$, Normal - $C,A$
Abnormal - $C$, Normal - $A,B$
Thus, this condition is satisfied in three outcomes, while only one has been counted. The probability for this case then becomes,
$$3\times\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}$$
Similarly, in the second case,
$$3\times\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{9}{64}$$
which gives the desired result.
In general, to get the number of outcomes for the particular case, we need to find out under how many combinations that case can occur. In the first case here, 1 child is abnormal, and can be selected from 3 children in $\binom{3}{1}=3$ ways. Thus, we have three such outcomes possible.
