Usually in any problem, whether or not I consider the order of arrangement, my answer is always the same. For example : suppose I have an urn with 2 white and 3 black balls. 2 black balls can be drawn in 6 ways (order taken into account) and any 2 balls can be drawn in 20 ways (order taken into account). So the probability that 2 balls drawn at random are both black is 3/10. Alternatively, 2 black balls can be drawn in 3 ways (order not taken into account) and any 2 balls in 10 ways (order not taken into account). So the probability is 3/10, which is consistent with the previous answer. Today I encountered a problem where the two answers differed.
$The$ $Problem :$ How many ways can 3 balls $(A,B,C)$ be distributed into 3 boxes $(I,II)$ such that box $I$ contains 1 ball?
Order of balls within each box not considered:
I list out all the possibilities - $ABC| $ ; $ |ABC$ ; $AB|C$ ; $BC|A$ ; $CA|B$ ; $A|BC$ ; $B|CA$ ; $C|AB$ ; $ |ABC.$
$2^3$ possibilities. Out of these 3 are favourable to our event. So the probability is 3/8, considering these are all equally likely.
Possibilities - $ABC| $ ; $ACB| $ ; $BCA| $ ; $BAC| $ ; $CAB| $ ; $CBA| $ ; $AB|C$ ; $BA|C$ ; $AC|B$ ; $CA|B$ ; $BC|A$ ; $CB|A$ ; $A|BC$ ; $A|CB$ ; $B|CA$ ; $B|AC$ ; $C|AB$ ; $C|BA$ ; $ |ABC$ ; $| ACB$ : $ |BCA$ ; $ |BAC$ ; $ |CAB$ ; $ |CBA.$
$[3+(2-1)]!$ possibilities. Out of these 6 are favourable to our event. So the probability is 1/4, considering the outcomes are equally likely. Which one is wrong and why? How should I be approaching a problem if in general, the two methods yield different answers?