Suppose we have an alien hand with F different fingers. We also have R different rings. How many ways there are to place those R rings on those F fingers considering that the order matters?
I can't see any mistake in the current approaches:
- Using one approach: Use stars and bars to get the possibilities of the amount of rings in the fingers and the multiply by $r!$ to place them in all possible orders.
- Using another approach: Let each ring select a finger. For the first ring, it has f possibilities. The second ring has f+1 possibilities, that is, f-1 fingers, and at the finger with a ring, it can be placed above or below... Using the same reasoning we get: $f\cdot (f+1)\cdot (f+2) \cdots (f+r-1) = \frac{(f+r-1)!}{(f-1)!}$
What is the correct approach in this case and why? Can someone please clarify it for me? It seems that both reasonings are correct, but of course it's not since the results are different...
Thanks!
Using the first approach, we have to put the bars in $f-1$ places among some $r+(f-1)$ linearly ordered places. There are $$ \binom{r+(f-1)}{f-1} $$ possibilities for this. For each choice of the bars we have $r!$ choices for the permutation of the rings to be arranged in order in the empty places. We get the same number, $$r!\binom{r+(f-1)}{f-1} $$ as in the second approach.