For the sake of argument let's say I have a set of fruits: ${(Apple, Banana, Grape, Pear, Orange)}$. In this scenario $n = 5$ (the size of the set of elements). For a set like this with distinct elements the following four formulae are used.
- Permutation, repetition: $n^r$ - total outcomes of picking a single fruit and returning it $r$ times and recording the result in an ordered list.
- Permutation, no repetition: $n!\over(n-r)!$ - total outcomes of picking a single fruit and placing it to the right side of the previous fruit it $r$ times. $r <= 5$, otherwise I run out of fruit, and imagine I am looking at the row of fruit at the end of the process [so order matters].
- Combination, no repetition: $n!\over(n-r)!r!$ - total outcomes from picking $r$ fruits and counting what I have when finished.
- Combination, repetition: $(r+n-1)!\over(n-1)!r!$ - imagine we have $r$ baskets of fruit, each with the same composition as the set described. This formula gives the outcomes from picking one fruit from each basket and counting what we have when finished.
My question is what happens when we add 3 bananas, 2 grapes and an orange to our original set. I think for the specific case of permutation with no repetition "to empty" [i.e. we are performing the task until we have no elements remaining] we can do the following:
Our set has $5 + 3 + 2 + 1 = 11$ elements in it. Total permutations without repetition is simply $11!$ We can account for duplicity by dividing by the factorial of repeat cases to yield $11!\over4!3!2!$ to account for the $4$ bananas, $3$ grapes and $2$ oranges we now have.
Are there similar techniques for other cases? For example, let us say the question was now find the combinations with no repetition of picking $7$ fruits from our $11$ fruit set. With a probability tree you can find the solution but are there better ways?
I am aware that the explanation may take some time to type up so even a link to a resource would be fantastic. Thanks in advance for any help!
I have seen this question: How to calculate combinations with duplicates? However, the answers are very specific to the specific question that was asked so I do not feel that this question is a duplicate of that. [I suppose though that the style of answers given there may be indicative of the fact that there aren't general ways to solve these questions].
P.S. Sorry for the wordy question - I feel like a lot of people struggle with when to use combination/permutation and repetition or not so I have tried to make my question as useful as possible to everyone that comes across it [at the expense of brevity it seems].