Recent content by zxen

That really doesn't help as to the uniqueness of ##3j+4k##. Even if we solve for the nonnegative solutions, some pairs of ##(j,k)## exist which will give the same value of ##3j+4k## and be clubbed into the same term.

Expanding the multinomial, the general term is 8!/i!j!k! * x^(3j+4k) for all i + j + k = 8. The number of terms would be the number of distinct powers of x, the number of distinct outputs of 3j+4k with the specified constraints for i, j and k. I attempted to make cases. 3j+4k where j+k <= 8...