I have stumbled across an interesting exercise regarding combinatorics and am trying to figure out whether I have been interpreting it correctly. The base of the problem goes as follows: “A lecture visited by 300 students at a university has a total of 10 different exercises groups (from A to J) which can be ranked by each student from 0 to 5, with 0 being “least preferred” and 5 being “best choice” for the student. Each student hast to rank every exercise group (before he will be accepted into one).”
The first task was about in how many different/ unique variations the exercise groups can be ranked, which I found to be n^k (since there are no restrictions)
As for task b) I am a bit unsure about my reasoning, the task says “Since some students found a trick on how to get quicker into a specific exercise group, by ranking a total of nine of the exercise groups with 0 and the one they wanted to attend with 5, there has been a restriction of having to rank at least 3 of the exercise groups with a positive ranking. In how many different ways can a student now rank the lecture’s exercise groups?”
As far as my understanding went; I considered the ranking of 3,4,5 to be the positive ranks, meaning a student could for instance request a rank of $${0 0 0 0 0 0 0 3 4 5} $$ or $$ {1 1 1 1 1 1 1 5 5 5} $$ etc.
Meaning he would have to choose a minimum of $$\binom{10}{3}$$ but could potentially choose more positive ranks as well. As soon as he would’ve chosen a specific/the minimum amount of positively ranked lectures the “empty spaces” could be filled with any other rank and ordered amongst themselves by $$(10-k)! $$
So my answer would therefore be, in order to find the total of all possible solutions for ranking all exercise groups by matching a minimum requirement:
$$\displaystyle \sum_{k=3}^{10}\binom{10}{k} * (10-k)!$$
Is it correct? I would be looking for possible thinking errors and/or corrections on my answer and even further approaches to come to a solution; I find combinatorics to be very interesting but sometimes am getting definitions mixed up in distinguishing particles of a problem. Any feedback would be of great help and further recommendations are greatly appreciated, thank you :) !
Originally you had $6^{10}$ possibilities. For similar reasons, if somebody gives positive scores (I feel "ranks" is the wrong word here) to $k$ exercise groups, you are going to have to multiply by $5^k$ for their possible scores (and by $1^{10-k}=1$ for those with $0$-score) in your later calculations.
Meanwhile I do not see why you have $ \times (10-k)!$. Your ${10 \choose k}$ is the number of ways of of choosing which $k$ to have positive scores and which $10-k$ to have $0$-scores. They are already in order A,B,...,J.
So $$\sum\limits_{k=3}^{10} {10 \choose k} 5^k$$ which should be the same as $$6^{10}-\sum\limits_{k=0}^2 {10 \choose k} 5^k$$ if you simply subtract prohibited scores.