In the binomial distribution, why does order not matter?

Question

(For context): $X$ is the random variable for number of successes ($x$) out of $n$ independent trials with probability of success $p$ and probability of failure $1-p = q$

The formula for mass my book gives me for the binomial distribution is this: $P(X=x) = {n \choose x}p^xq^{n-x}$

My question is this. How come order doesn't matter? Why do we use ${n \choose x}$ in the above to calculate all the ways we can get $x$ successes out of $n$ independent trials instead of using $nPx$? I.E. Why do we use combination instead of permutation? Can someone explain this to me because it's been really bothering me.

Thank You!

Answer 1

Look at a small example. In a sequence of $4$ trials, what are the ways to get $2$ successes? With $S$ for success and $F$ for failure they are $SSFF$, $SFSF$, $SFFS$, $FSSF$, $FSFS$, AND $FFSS$; there aren’t any more. Specifically, there is one way to get $2$ successes for every pair of trials, and there are $\binom42$ pairs of trials.

The successes and failures don’t have individual identities: their only identities are as success or failure and where in the sequence of trials they occur. You might as well imagine that you have $x$ white and $n-x$ black stones, identical in all respects save color, and want to know in how many ways you can line them up in a row. The stones aren’t labelled; all that matters is the color of each stone and its position in the row. For instance, there is only one distinguishable arrangement with all $x$ of the white stones together at the head of the line.

It would be another story if the white stones were labelled $1$ through $x$: then there would be $k$ distinguishable arrangements with all of white stones together at the head of the line. That’s the model that your proposed calculation would be following, and it clearly doesn’t fit what you’re trying to model: the individual successes don’t come with individual labels beyond success in trial number so-and-so.

Answer 2

If we have three events $A,B,C$ where $A,B$ are successes and $C$ is a failure, then $A,B,C$ and $B,A,C$ both represent two successes in the first two trials and a failure in the last trial, so they are considered to be the same.

Answer 3

This is tautological, or definitional. You only use the binomial distribution when the order of the successes doesn’t matter because the successes are all interchangeable and independent. If the order of the successes matters, or the successes aren’t interchangeable, or they’re not independent, then you shouldn’t be using the binomial distribution, lol.