inclusion and exclusion practice exercise

Question

This is a practice question. I understand Inclusion and Exclusion but I am having a terrible time setting questions up correctly.

This is on of the practice questions in the text.

At a flower shop they want to arrange $15$ different plants on five shelves for a window display. In how many ways can they arrange them so that each shelf has at least one, but no more than four plants ?

My understanding is that the equation would be of the form $x_1+x_2.....+x_5=15$ where $1 \le x_i \le 4$ but then we modify it so that it is $y_1+y_2+y_3...+y_5=15-5$ where $0 \le y_i \le 4$

$|S|= \binom{5+10-1}{10}$ and $N(c_i)= \binom{5+5-1}{5}$ where $1 \le i \le 5 $ so $|S_1|=5* \binom{5+5-1}{5}$

Am I setting this up correctly. Please note I am very new to the subject of inclusion and exclusion.

Answer 1

So had to go through the basics again but I figured it out. The answer is

$15![(\binom{5+10-1}{10})-(\binom{5}{1})(\binom{5+6-1}{6})+(\binom{5}{2})(\binom{5+2-1}{2})]$

main Principles used are Inclusion and exclusion, Combination with Repetition.

Answer 2

The way inclusion-exclusion can be used here is: for each subset $S$ of $\{1,2,3,4,5\}$ (representing the shelves) let $N_S$ be the number of ways to do the arrangement if the shelves in set $S$ are required to be empty (the other shelves may or may not be empty). Write the number of arrangements where at least one shelf is empty in terms of these, using the Inclusion-Exclusion formula.
Subtract from the total number of arrangements, and you have the number where no shelf is empty.