Factoring Factorial Fate

The only collection of four dice rolls that has a product of $6!$ is $(6,6,5,4)$ . Thus $P_4 \; = \; \frac{1}{6^4} \times \binom{4}{2} \times 2 \; =\; \frac{1}{108}$ There are four different sets of five dice rolls that multiply to $6!$ , namely $(1, 4, 5, 6, 6)$ , $(2,3,4,5,6)$ , $(2, 2, 5, 6, 6)$ and $(3, 3, 4, 4, 5)$ . Note that the first possibility has one repeated digit, the second possibility has no repeated digits, while the third and fourth have two repeated digits. Thus $P_5 \; = \; \frac{1}{6^5}\left[\binom{5}{2} \times 3! \;\;+\;\; 5! \;\;+\;\; 2 \times 5 \times \binom{4}{2}\right] \; =\; \frac{5}{162}$ There are six different sets of six dice rolls that multiply to $6!$ , namely $(1, 1, 4, 5, 6, 6)$ , $(1, 2, 2, 5, 6, 6)$ , $(2, 2, 3, 3, 4, 5)$ , $(1, 3, 3, 4, 4, 5)$ , $(1, 2, 3, 4, 5, 6)$ and $(2, 2, 2, 3, 5, 6)$ . Note that the first four possibilities have two repeated digits, the fifth has no repeated digits, while the last has one digit repeated three times. Thus $P_6 \; =\; \frac{1}{6^6}\left[4 \times \binom{6}{2}\times\binom{4}{2}\times2 \;\;+\;\; 6! \;\;+\;\; \binom{6}{3}\times3!\right] \; = \; \frac{65}{1944}$ and hence $\boxed{P_4 \; < \; P_5 \; < \; P_6 }$