There is no need to remember any formulas.
This is a question of Discrete Maths.
Consider this simple question:
Q: How many pairs of single-digit positive integers (A, B) are there such that A > B.
Solution:
A and B can have values between 1 and 9 (both inclusive).
For A = 1, no value of B satisfies the given condition.
For A = 2, we can have B = 1, i.e one value.
For A = 3, we can have B = 1, 2, i.e 2 values.
...
For A = 9, we can have B = 1, 2, ..., 8, i.e. 8 values.
You can see that summing these will give the answer. This is the summation of first 8 natural numbers.
Why we get 8? Because A can have any one of the 9 values, where the smallest value makes no value available for B.
Another approach:
Total number of ordered pairs possible (A, B) = $n^2$
Out of these $n^2$ pairs, $n$ pairs will be of equal numbers, i.e. A = B.
when n = 9, we have total pairs = $9^2$, out of these pairs, $9$ pairs are when both number are equal. Like (1, 1), (2, 2), ..., (9, 9).
So, now we are left with $n^2 - n$ pairs of unequal numbers. Out of these exactly half will be such that A > B and another half will be A < B. So, this gives us $(n^2 - n)/2$, or $n(n-1)/2$, i.e. summation of first $n-1$ natural numbers.
Now getting to your question, Suppose we are given $n$ bits. Total possibilities for any number = $2^n$. From the above argument, total number of possible pairs such that A > B is the summation of $2^n - 1$.
$\ \ = \frac {(2^n - 1)2^n}{2}$
$\ \ = \frac {2^{2n} - 2^n}{2}$
This gives your formula.
Answer is: $\ \ = \frac {2^{8} - 2^4}{2}$ = 120