I have to evaluate the following quantity
$P\left(C|AB\right)$
Where A, B, and C are events.
Since I know the following conditional rule
$P\left(A|B\right)=\frac{P\left(AB\right)}{P\left(B\right)}$
I thought I could write
$P\left(C|AB\right)=\frac{P\left(CAB\right)}{P\left(AB\right)}$
and
$P\left(C|AB\right)=\frac{P\left(BC|A\right)}{P\left(B|A\right)}$
However, only the latter equation gives me the correct result. Is the other one incorrect?
Both equations are correct as stated by @geetha290krm as we will demonstrate. The conditioning rule holds that for events $X, Y$ having $P(Y) \neq 0$, $$P(X \vert Y)=\frac{P(X \cap Y)}{P(Y)} \tag{1}.$$ We would like to show that $$P(C \vert A \cap B) = \frac{P(A \cap B \cap C)}{P(A \cap B)} = \frac{P(B \cap C \vert A)}{P(B \vert A)} \tag{2}.$$ Applying $(1)$ with $X = C, Y= A \cap B$, we immediately obtain the first equivalence in $(2)$, assuming $P(A \cap B) \neq 0$. Notice that this implies that $P(A) \neq 0$. We can therefore proceed by multiplying the right hand side of the first equivalence by $1$ in the form $\frac{\frac{1}{P(A)}}{\frac{1}{P(A)}}$ to find that $$P(C \vert A \cap B) = \frac{\frac{P(A \cap B \cap C)} {P(A)}}{\frac{P(A \cap B)}{P(A)}} = \frac{\frac{P((B \cap C) \cap A)} {P(A)}}{\frac{P(B \cap A)}{P(A)}} = \frac{P(B \cap C \vert A)}{P(B \vert A)},$$ recognizing conditional probability $P(X \vert Y)$ with $X= B \cap C, Y= A$ in the numerator and $X=B, Y=A$ in the denominator. This complete the proof of $(2)$.