How many different basis' exist for an n-dimensional vector space in mod 2? Imagine a 2-dimensional vector space in $\mathbb{Z} /2 \mathbb{Z}$. The only possible basis' are
$$ 
\left(\begin{array}{c}
1\\
0 \end{array}\right), 
\left(\begin{array}{c}
0\\
1 \end{array}\right); \left(\begin{array}{c}
1\\
1 \end{array}\right), 
\left(\begin{array}{c}
0\\
1 \end{array}\right);\left(\begin{array}{c}
1\\
0 \end{array}\right), 
\left(\begin{array}{c}
1\\
1 \end{array}\right)$$
This means for n = 2 would 3 be the correct answer. For n = 1 the answer is 1.
What is the correct answer for a variable n?

We can do the more general case for $F = \mathbb{F}_q$ for $q$ a power of a prime. The first vector can be chosen from $q^n - 1$ possible choices. The subspace generated by it has $q$ points, since each vector is determined by $q$ possible coefficients. Continuing we get the formula $(q^n - 1)(q^n - q)\cdots (q^n - q^{n-1})$. This is the number of ordered bases though, but we can just divide by $n!$ to get the number of unordered bases.
For instance in your case $q = 2^1 = 2$ and $n = 2$, so we get $(2^2-1)(2^2-2) = 3\times 2 = 6$ possibilities. To get your $3$ we just divide by $2!$. You could write the general formula to be
$(q^n - 1)(q^n - q)\cdots (q^n - q^{n-1})/n!$
