I'm trying to figure out how many possibilities are possible under these conditions:
I got 32 items (marbles) and need to place them into 8 groups where there's at least 1 marble in each group.
Also groups are labeled so the order matters: $[25,1,1,1,1,1,1,1]$ and $[1,1,1,1,1,1,1,25]$ are different.
Also what if a group could only have maximum of $8$ items.
How would anyone solve these? I don't even know where to start. Thank you for your help.
(a) Use the stars and bars method.
Put $\sum_{i=1}^8x_i=32$. As $x_i\geq1$, we can write $\sum_{i=1}^8y_i=24$ where $0\leq y_i\leq7$.
So, the final answer should be $$\binom{24+8-1}{8-1}$$
(b) Use the stars and bars method along with the inclusion-exclusion principle.
Put $\sum_{i=1}^8x_i=32$. As $x_i\geq1$, we can write $\sum_{i=1}^8y_i=24$ where $0\leq y_i\leq7$.
Using the inclusion-exclusion principle, the final answer should be
$$\binom{24+8-1}{8-1}-\binom{8}{1}\binom{16+8-1}{8-1}+\binom{8}{2}\binom{8+8-1}{8-1}-\binom{8}{3}\binom{0+8-1}{8-1}$$
The second term arises by taking the case where $1$ box has over $8$ objects, the third term arises by taking the case where $2$ boxes have over $8$ objects, and the fourth term arises by taking the case where $3$ cases have over $8$ objects.
