Three fair dice are rolled. What is the probability that the sum of the three outcomes is 10 given that the three dice show different outcomes?

Question

I know how to solve this question using conditional probability. I tried using another method of solving it which gives me another answer(which i know is wrong), help me find the fault in it.

The $3$ cases where the sum of $3$ distinct outcomes of the die is $10$ are: $A=(3,5,2)$ $B=(4,5,1)$ $C=(1,3,6)$. Probability $(A) = 3!\cdot (1/6)^3$ .

Reasoning.

$(1/6)^3$=probability to get $3/5/2$ (they are independent events)

$3!$ = number of ways to arrange $3,5,2$

This is the same for $B,C$ as well. Hence the final answer should be $3\cdot p(A)=1/12$. The actual answer is $3/20$ (solve using conditional probability). What are the wrong steps I have assumed or taken?

Answer 1

Your mistake is that once you're given that the dice show different outcomes, then $3/5/2$ are no longer independent events. If the first die is $3$, neither of the other two dice can be.

Answer 2

You should divide the number of favourable cases, $3$ (that is $(2,3,5),(1,4,5),(1,3,6)$), by the number of possible distinct outcomes for three dice, $\binom{6}{3}$: $$\frac{3}{\binom{6}{3}}=\frac{3}{20}.$$ or, equivalently, divide $3\cdot 3!$ by $6\cdot 5\cdot 4$: $$\frac{3\cdot 3!}{6\cdot 5\cdot 4}=\frac{3}{20}.$$