Given $A_1, A_2,\ldots A_n$ be i.i.d random variables $\sim$ an arbitrary distribution on $\left\{1,2,3\right\}$. Define $P = $ family of discrete distributions on $\left\{1,2,3\right\}$. Find a sufficient statistics for $P$.
My attempt: We define $T(\vec{A}) = A_1A_2\ldots A_n$. Let $Q\in P$, and define $Q(A_i = 1) = p_1$, $Q(A_i = 2) = p_2$ and $Q(A_i = 3) = p_3$ with $\sum_{i=1}^{3} p_i = 1$
Now, we would show: $P(\vec{A}= (a_1, a_2,\ldots, a_n)|T(\vec{A}) = t)$ is independent of $p_i$ for $i\in \left\{1,2,3\right\}$. But this is true because for any $t$ such that $T(\vec{A}=t)$, $t$ would have the form $1^{n-k-m}2^{k}3^{m}$ for some non-negative integers $k,m\leq n$ (by Unique Prime Factorization Theorem).
Therefore, $P(\vec{A}= (a_1, a_2,\ldots, a_n)|T(\vec{A}) = t) = \frac{n!p_2^k\ p_3^m\ p_1^{n-k-m}}{{n \choose k} {n-k \choose m}p_2^k\ p_3^m\ p_1^{n-k-m}} = \frac{n!}{{n \choose k} {n-k \choose m}}$ (is this formula correct?), which is independent of $p_1, p_2, p_3$ (1)
If no such $t$ exists for a given $\vec{A}$ where $T(\vec{A})=t$, then $P(\vec{A}= (a_1, a_2,\ldots, a_n)|T(\vec{A}) = t) = 0$, which is triviallly independent of $p_i$s (2)
From (1) and (2), $T(\vec{A}) = A_1A_2\ldots A_n$ is a sufficient statistics for $P$.
My question: Could anyone please help verify if my solution above is correct, particularly the formula for $P(\vec{A}= (a_1, a_2,\ldots, a_n)|T(\vec{A}) = t)$? Any thought/help would be really appreciated.