Combinatorics for number of distinct terms in multinomial expansion

[zxen]

Homework Statement

If the number of terms in the expansion of (1+x^3 + x^4) ^8 is N, find the difference in the digits of N.

Relevant Equations

Multinomial expansion.

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 would have maximum 45 solutions. (Using stars and bars for solving j + k = 8 - i)

But this overcounts, because of cases where j = 0,4,8 and k = 0,3,6 (multiples of 4 and 3 resp.) 3j+4k has a repeated value.

Is there some way I can obtain the answer without actually having to count values of 3j+4k for j=0,4,8 and k=0,3,6?
For reference, there's this answer on AOPS which I couldn't understand. https://artofproblemsolving.com/community/c4h1789008p11825444

 

[WWGD]

Just a thought: How about breaking it into $((a+b)+c)^8$, knowing that $(a+b)^8$ has$9$terms?
Edit: The other way is counting the nonnegative solutions to $x_1+x_2+x_3=8$.
Edit 2: Ouch, it seems these two don't agree with each other. Let me double-check.

 

[zxen]

Just a thought: How about breaking it into $((a+b)+c)^8$, knowing that $(a+b)^8$ has$9$terms?
Edit: The other way is counting the nonnegative solutions to $x_1+x_2+x_3=8$.
Edit 2: Ouch, it seems these two don't agree with each other. Let me double-check.

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.