Here is the problem: An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 ace.
My solution:
$$\frac { {4 \choose 1} {48 \choose 12} {3 \choose 1} {36 \choose 12} {2 \choose 1} {24 \choose 12} } {{52 \choose 13} {39 \choose 13} {26 \choose 13} }.$$
I'm not sure if I did it right because the book gives the following solution.
Solution from the book:
E1 = {the ace of spades is in any one of the piles}.
E2 = {the ace of spades and the ace of hearts are in different piles}.
E3 = {the aces of spades, hearts, and diamonds are all in different piles}.
E4 = {all 4 aces are in different piles}.
The desired probability is $P(E1E2E3E4)$, and by the multiplication rule, $P(E1E2E3E4) = P(E1)P(E2|E1)P(E3|E1E2)P(E4|E1E2E3)$:
$$ P(E1 E2 E3 E4) = \frac {39 \cdot 26 \cdot 13} {51 \cdot 50 \cdot 49} .$$
I do understand that the book gave a neat solution, but I wonder if my logic was correct for this problem, because I did not realize how to use the multiplication rule here.
