The following question was posed in my Mathematical Statistics book amounting to the following:
"3 words describe 3 pictures. Assuming you assign the words to the pictures randomly, give the probability distribution for Y (the number of correct matches)."
To try to solve the problem, I did
$$\frac{3 \choose Y}{3!}$$.
This gave me the correct answer for $P(3)$ and $P(1)$ and I just did $1-(P(3)+P(1))$ to get $P(0)=\frac{1}{3}$ (note that $P(2)=0$ as two correct matches implies a third). For kicks though, when I tried to apply the "choose" strategy instead to $Y=0$, I get $\frac{1}{6}$ for the probability rather than the correct answer of $\frac{1}{3}$. Why doesn't the 3 choose y divided by 3! method work for $Y=0$? Why does this breakdown on $Y=0$? Thinking about this more, I realize the needing for none of them to be in the right place would do restrict the ordering, but I can't seem to articulate rigorously why.
The generalized formula for if you have a total of $n$ pictures and words and you are asking the probability of getting exactly $Y$ correct matches when randomly assigning the pictures to the words would be:
$$\dfrac{\binom{n}{Y}\cdot !(n-Y)}{n!}$$
where $!k$ represents the subfactorial function, returning the number of derangements on a set of size $k$.
The explanation being, we first select which of the pictures were correctly matched with their respective words. We then pick how to mix up the remaining pictures and words such that no picture gets its correct respective word (exactly the scenario that the subfactorial counts). We finally divide by the number of ways of assigning the words to the pictures in the first place.
Indeed, with this correction your formula now works for all inputs: $Pr(Y=0)=\frac{!3\binom{3}{0}}{3!}=\frac{1}{3},Pr(Y=1)=\frac{!2\binom{3}{1}}{3!}=\frac{1}{2}$
$Pr(Y=2)=\frac{!1\binom{3}{2}}{3!}=0,Pr(Y=3)=\frac{!0\binom{3}{3}}{3!}=\frac{1}{6}$