Imagine the dice as being tossed one at a time, or imagine them different-coloured. Then there are, as you point out, $6^5$ outcomes. They are all equally likely.
So for example the outcome $66666$ is just as likely as the outcome $12345$, which in turn is just as likely as $54321$.
But for the game, the order does not matter. Thus a reaasonable-sounding way to list the outcomes is just to list the results on the dice, say in non-decreasing order. Thus $12345$ is one outcome, and $22346$ is another, and $16666$ still another. But we do not list $66166$, since that's the "same" outcome as $16666$.
In principle, that's fine, we can define as we wish what we mean by outcome. The big problem is that if we do not consider order, then not all outcomes are equally likely. For example, five $1$'s is substantially less likely than getting a pattern of $5$ numbers that consists of the numbers $1$, $2$, $3$, $4$, and $5$. Indeed, the latter is $120$ times as likely as the former.
Thus if we use sequences as outcomes, or equivalently consider the dice to be of different colours, we get that all outcomes are equally likely, and we can obtain other probabilities by simply counting.
If we decide instead to not distinguish between the dice, then all outcomes are no longer equally likely, and computing probabilities gets more complicated. That's why Yahtzee books recommend using the $6^5$-element sample space. We can indeed work with a smaller sample space, but then analysis of probabilities becomes more difficult.
Using the $6^5$-element sample space, we can see fairly easily that four $6$'s and a $3$ can happen in $5$ of our $6^5$ equally likely outcomes, so has probability $\dfrac{5}{6^5}$. And we can see that getting the numbers $1$, $2$, $3$, $4$, and $5$ showing on our dice happens in $5!=120$ of our outcomes, so has probability $\dfrac{120}{6^5}$.
Best Answer
** How many outcomes are in the event where nobody rolls a six? If they can't roll a six, there are 5 other numbers to roll, and either coin-flip is still allowed. So each person has $2\times 5=10$ possible outcomes. Since there are 5 people, there are $10^5$ possible outcomes.
** How many outcomes are in the event where at least one person rolls a six? Well, either there are no sixes rolled or at least one person rolls a six, so this result is the difference of your answer and the answer above $12^5 - 10^5$.