Identifying combinations removing 2 possiblities from a set If I have a set A,B,C,D,E,F,G

I am looking for a recipe to identify all possible combinations of letters removing all combinations of 2 letters. 
In this case the result would be 
CDEFG remove AB
ABEFG remove CD
ABCDG remove EF
BDEFG remove AC
ACEFG remove BD

etc..

I want to know all the removal possibilities
 A: Number of combinations =$5!\times^7C_5$
A: The number of ways to choose the two that you remove is the number of ways to choose $2$ objects from a set of $7$. This is known as a combination, and is written and calculated as follows:
$$\binom{7}{2} = \frac{7!}{(7-2)!\,2!} =-\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{\left(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\right)\left(2 \cdot 1\right)} = \boxed{21}$$
The exclamation marks here stand for factorials.

Instead of choosing the two that you remove, you could also choose the $5$ that you don't remove. This would be $$\binom{7}{5}$$ which evaluates to the same number. In general, $$\binom{n}{k} = \binom{n}{n-k}$$

A list of all $21$ possibilities for the two letters removed: $$\mathrm{AB, AC, AD, AE, AF, AG, BC, BD, BE, BF, BG,} $$
$$ \mathrm{CD, CE, CF, CG, DE, DF, DG, EF, EG, FG}$$
A: So you want, not just the number of such possibilities but the actual lists themselves?  The simplest way to do that is to list all combinations in a specific order.  Taking "A" as the first letter they are "AB", "AC", "AD", "AE", and "AF".  Now B- "BC", "BD", "BE", "BF" (we do not count "BA" since that was already given as "AB".  We list each letter with all succeeding letters.).
"CD", "CE", "CF".  "DE", "DF".  "EF".  Now it is easy to list the 5 letters left after we remove those.  Notice that there are a total of 6+ 5+ 4+ 3+ 2+ 1= 21 as others said.
A: It seems like you are looking for an algorithm? It seems that your problem is equivalent to finding all the pairs of letters in the set. The below algorithm will generate the pairs you are looking for where $S$ is a set of $n$ letters. The following Python code will generate the pairs you are looking for.
(I tried to write it in pseudocode, but I am not familiar with mechanisms on StackExchange to do that nicely).
def findPairs(S): #given a set S, return all the pairs
    pairs = []
    for i in range(0,len(S)-1): #for each letter
        for j in range(i+1,len(S)): #for each letter past the first one we chose
            pairs.append(S[i]+S[j]) #add the pair of letters, '+' is string concatenation
    return pairs

Notice that in this algorithm, we avoid duplicates like $AB$ and $BA$ by only picking pairs where the index of the first letter is less than the index of the second.
