Wait: re-examining — if the *profile* is unordered, and we are to count distinct multisets of eruption levels, then yes, number of combinations is number of solutions to $x_1 + x_2 + x_3 = 4$, $x_i \geq 0$, with $x_L, x_M, x_H$ counts: $\binom{6}{2} = 15$.

Sat, 09 May 2026 04:45:52 +0000

Content under review