I'm really weak when it comes to discrete math and combinatorics, so I wanna make sure whether I'm thinking corectly.
An ice cream store sells 5 different flavours of ice cream. For a bowl, one chooses at random three kinds (not necessarily different); the order does not matter. In how many ways can 6 people buy such bowls so that that each flavour is taken at least once.
I recognize this as inclusion-exclusion principle. Let's assume we have $n$ different, numbered flavours. Using "stars and bars" we find
$${3+n-1\choose n-1}$$ ways of making arbitrary bowls from $n$ flavours.
Now lets setup the inclusion-exclusion principle:
- $X$ - 6 people buy arbitrary bowls: $${3+5-1\choose 5-1}^6$$
- $A_i$ - nobody bought $i$th flavour, 4 left: $${3+4-1\choose 4-1}^6$$
- $A_i\cap A_j$ - nobody bought $i$th nor $j$th flavour, 3 left: $${3+3-1\choose 3-1}^6$$ and so on. So the answer, according to this logic, should be $$\sum_{i=0}^4(-1)^i{5\choose i}{3+(5-i)-1\choose (5-i)-1}^6=\sum_{i=0}^4(-1)^i{5\choose i}{7-i\choose 4-i}^6$$ Is that correct? Thanks.
If each person is choosing uniformly and at random, (remember, a person is allowed to choose all $3$ of the same flavor), we want yo count putting $18$ distinct balls being put in $5$ distinct boxes with no box empty.
So either we count it as $\Large5!{18\brace k}$, using Stirling numbers of the second kind, or
use inclusion-exclusion, $\Large 5^{18} - \binom51 4^{18}+\binom52 3^{18}-\binom53 2^{18} +\binom541^{18}$