Counters are kind of like coins with two different colours on each side. By distinguishable I mean that each of the 6 counters has different colours. For instance, 1= blue and green, 2= red and black, 3= silver and golden, and so on..
So how many outcomes are possible if the counters are distinguishable.
Take each counter one by one. Now each one of them has two possibilities irrespective of the outcomes of the other counters. In short each counter has results which are independent of the other counters.
So let's say if there was only one counter. Then the answer would simply be $2$. If there were two counters, for each outcome of the first, we have two of the second. Thereby giving us
$$N = 2*2 = 4$$
Extrapolating this philosophy for $6$ counters, we have
$$N = 2*2*2*2*2*2 = 2^6 = 64$$
So number of outcomes is $64$. Keep in mind that this is the total number of possibiliyies if you fix the order of flipping i.e. say that Blue/Green is flipped $1st$, Red/Black is flipped $2nd$ and so on so forth. If the order is not fixed either, you multiply this with $6!$ (number of permutations of counters), you get
$$N = 64*6! = 46080$$
If the counters are indistinguishable, this becomes a multiple coin toss problem with no permutations possible as coins are all same and total no of outcomes as
$$N = 64$$
I am also attaching a link of a basic coin toss guide
http://www.probabilityformula.org/coin-toss-probability.html