Confusion regarding multiple events in bayes theorem I have gone through the chain rule of probability
Which specified
$P(A1,A2…An)=P(A1)P(A2|A1)…P(An|A1,A2…An−1)$
But when I see the derivation of Bayes rule for three events
I came across the formula to be
Source: 
$P(ABC) \;= P(A|BC)P(BC)\\
\qquad\quad\quad= P(B|AC)P(AC)\\
\qquad\quad\quad= P(C|AB)P(AB)$
But if we see according to the chain rule
$P(ABC) = P(A)P(B|A)P(C|A,B)$
So why the two are different is there a mistake in my understanding
Kindly enlighten me
Regards,
Siddartha C.S
 A: They are all true, so there is no inconsistency.
More generally by symmetry
$\quad P(A,B,C) \\
= P(A\mid B,C)P(B,C) = P(A\mid B,C)P(B\mid C)P(C)= P(A\mid B,C)P(C\mid B)P(B)\\
= P(B\mid A,C)P(A,C) = P(B\mid A,C)P(A\mid C)P(C)= P(B\mid A,C)P(C\mid A)P(A)\\
= P(C\mid A,B)P(A,B) = P(C\mid A,B)P(A\mid B)P(B)= P(C\mid A,B)P(B\mid A)P(A)\\
$
where you can see your first block of equalities in the left column and your chain rule equalities at the start
A: $$P(A|BC)P(BC) = \frac{P(ABC)}{P(BC)}P(BC)=P(ABC)$$
$$P(B|AC)P(AC) = \frac{P(ABC)}{P(AC)}P(AC)=P(ABC)$$
$$P(C|AB)P(AB) = \frac{P(ABC)}{P(AB)}P(AB)=P(ABC)$$
A: They are equivalent.   It is just varying the degree to which is applied the definition of conditional probability.
$\def\P{\operatorname{\mathsf P}}\P(A_1,A_2,A_3) ~{= \P(A_1)\P(A_2,A_3\mid A_1)\\= \P(A_1)\P(A_2\mid A_1)\P(A_3\mid A_1,A_2)
\\ = \P(A_1,A_2)\P(A_3\mid A_1,A_2) 
\\ ~\vdots~\textit{et cetera}}$
Using the "product" notation
$\P(ABC) ~{= \P(A)\P(BC\mid A)
\\ = \P(A)\P(B\mid A)\P(C\mid AB)
\\ = \P(AB)\P(C\mid AB)
\\ ~\vdots~\textit{et cetera}}$
