Permutation of 7 rings on 5 fingers

Question

If we have 7 rings of different colours and 5 fingers, then what's the number of ways can we wear them? The answer given in the book is $\frac{11!}{4!}$.

The answer is not $7^5$ or $5^7$. They are considering the arrangement of rings in each finger also.

Answer 1

At least one ring should be held by the person so u take extra 6 fingers just for reference not real , then no of ways of selecting 7 fingers is $11C_7$ U can see u are at least selecting 1 real finger , now as the rings are different u have arrange them so multiply by 7! Problem solved if u have any doubts u can comment

Answer 2

Consider the fingers as 5 boxes, and rings are distinctive balls, and orders matter(this is why we cannot use the $5^7$). So
|O(red)O(green)|O(yellow)O(black)|O(pink)|O(white)|O(orange)|
and
|O(green)O(red)|O(yellow)O(black)|O(pink)|O(white)|O(orange)|
are different cases.

This problem could be reduced to: There is 11 positions, 7 for the balls and 4 for the identical walls(the walls in the ends cannot be moved) That means we need to find 4 position to place the walls -- ${11 \choose 4}$ We need to permute the 7 distinctive balls to place them in the remain positions. -- $7!$

So together, it would be ${11 \choose 4}*7! = \frac{11!}{4!}$.

The trick here is the fingers are stacks, so the order of rings on the finger matters.