Consider the 9 qubit Shor code. This can detect and correct arbitrary single qubit errors, but if there are 2 or more single qubit errors before a correction round, the correction will fail. (In the sense that it won't reproduce the original logical state.) Hence the probability of failure of the Shor code is the probability of there being 2 or more single qubit errors. If we assume that single qubit errors occur independently with probability $p$, then we can calculate this probability as
$$ P(\mathrm{failure}) = 1 - \left[(1-p)^{9} + 9p(1-p)^{8}\right]. $$
We can write this probability as a series in $p$ as
$$ P(\mathrm{failure}) = \sum_{m=2}^{9} (-1)^{m} (m-1) \binom{9}{m} p^{m}. $$
My question is about the intuition for the coefficients of each term in this sum.
The $\binom{9}{m} p^{m}$ part seems natural. We can interpret it as the probability that $m$ single qubit errors occur multiplied by the number of groups of $m$ qubits these errors could act on.
Question: What I am confused by is the $(-1)^{m} (m-1)$ part of the coefficient. Why should some of the higher order terms lead to a suppression of the failure probability, and why should the weighting of these terms beyond the binomial coefficient increase with order? Is there some intuition for these coefficients?
Note 1: I realise that the $m=2$ case is the most important, since we typically assume $p \ll 1$. Indeed, most treatments truncate the series at $m=2$, where $(-1)^{m} (m-1) = 1$, so even if the higher-order terms do have an effect, that effect should be fairly negligible.
Note 2: Though I have specified the Shor code for concreteness, this behaviour is not specific to the Shor code. For example, if we restrict ourselves to bit-flip errors, then the failure probability of the bit-flip code will exhibit similar coefficients.