17. Carla rolls, all together, 4 dice, with faces numbered from 1 to 6. What is the probability that the product of the 4 numbers rolled is 24?
(A) $5 / 162$
(B) $7 / 324$
(C) $1 / 36$
(D) $13 / 324$
(E) $5 / 144$
(17) The correct answer is $(D)$.

Given that $24=2 \cdot 2 \cdot 2 \cdot 3$, let's see which scores can be obtained from the different dice so that their product is 24 (without considering the order):

- case $a$: they can be $2,2,2,3$;
- case $b$: they can be $1,2,4,3$;
- case $c$: they can be $1,2,2,6$;
- case $d$: they can be $1,1,4,6$.

Let's count how many outcomes of the rolls fall into each of the 4 groups of outcomes listed above. In the count, it will be useful to consider the order of the scores, for example, imagining that the dice are of 4 different colors and considering two outcomes equal if and only if the same scores appear in the same colors.

In group $a$, there are 4 outcomes (it involves choosing which die shows the 3).

In group $b$, there are $4 \cdot 3 \cdot 2=24$ outcomes (it involves choosing which of the 4 scores 1, 2, 3, 4 appears on the first die, which of the 3 remaining scores appears on the second, and which of the 2 remaining scores appears on the third).

In group $c$, there are $4 \cdot 3=12$ outcomes (it involves assigning one of the 4 dice to the score 6 and one of the 3 remaining dice to the score 1).

In group $d$, there are 12 outcomes, as in $c$.

The favorable outcomes for the event in question are therefore $4+24+12+12=52$.

The required probability is thus $\frac{52}{6^{4}}=\frac{13}{324}$.

Question proposed by Paolo Francini.