Number of possible combinations of pairs (order important) I'm not very sure how to go around explaining this, so let me present an example of a smaller scale of 2 items, assuming that the pairs of those two chosen items are A/1 and B/2. The possible combinations would be:
{AB, 1B, A2, 12, BA, B1, 2A, 21}

Thus, the amount of combinations would be 8, in this case. How much would it be if the scale was 6 instead of 2 items? Assuming that the pairs are A/1, B/2, C/3, D/4, E/5, and F/6, as well as the order mattering.
Possible combinations would include:
{ABCDEF, 1BCDEF, BD5FA3, FEDCBA, 654321, 1B53F4}

How would I go about calculating this?
 A: I don't know your level of maths, so feel free to skip any parts you know already.
The counting principle comes in handy here:

Suppose that two experiments are to be performed. Then if experiment 1 can result
in any one of $m$ possible outcomes and if, for each outcome of experiment 1, there
are $n$ possible outcomes of experiment 2, then together there are $mn$ possible outcomes of the two experiments.

You can think of an experiment as something that has at least one outcome. For instance, flipping a coin has 2 outcomes and rolling a 6-sided die has 6 outcomes. If we flip a coin AND  roll a die, we have $2\cdot 6$ outcomes because each outcome from the coin flip is associated with 6 possible outcomes of the die. The outcomes are $\{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6\} \to 12\textrm{ outcomes}$ where $H,T$ mean heads and tails, respectively.
There are 2 parts to this. The first part is to choose which element we choose from the pair ($A$ or $1$, $B$ or $2$,...). For each pair there are 2 choices. The second part is permuting the string of length 6 (2 in the first example), which just means arranging them in an order. For example, permuting 3 items has 6 outcomes: $\{ABC,ACB,BAC,BCA,CAB,CBA\}$
In the first part, we have 6 experiments, one for each pair, so by the principle of counting, we have $2^{6}$ outcomes.
In the second part, we have to permute 6 items. This is tantamount to choosing a first item, then a second item, then ..., then a sixth item. For the first item we have $6$ outcomes, for the second item we have $5$ outcomes because one of the items was already chosen to be first, ..., for the sixth item there is just $1$ choice. By the principle of counting there are $6\cdot 5 \cdot ... \cdot 1=6!$ outcomes.
Then, we apply principle of counting to the whole expreriment, yielding $2^{6}\cdot 6!$ outcomes, also mentioned by user2661923 in the comments.
For the first example, the analogous result is $2^{2}\cdot 2!=8$, which agrees with your answer.
