Let’s suppose we have $k$ sticks and we want to put them between $n$ circles where $n$ and $k$ are $2$ natural numbers.
In how many ways can we put the sticks between the circles?
Please note that the conditions are:
(I’m using an example with $2$ sticks and $5$ circles:) $1 \cdot$The $k$ sticks are identical.For example if you put two sticks between the first and the second circle, you can’t make a new combination by switching them.
$2 \cdot$You can’t change the position of the circles.Note that the circles are not identical so if you put the two sticks between the first and the second circle , you can make a new combination by putting the two sticks between the fourth and the fifth circles.
$3 \cdot$You can put the sticks wherever you want.(They can also be put next to each other with no circles between them. Also, you can put them outside the circles so that means that there can be only one circle to the right/left of a stick.
Here’s what I’ve tried: I thought since there are $k$ sticks, that means that the sticks make $k+1$ sections, where there could be $0$ to $n$ circles in each of the sections.
So: If we find how many $k+1$ non-negative integers there are such that their sum is $n$ , we’ll find the answer.
Since nobody has answered , I’m going to answer my question using @PM2Ring explanation: So first , we assume that the sticks are not identical and we can change the position of the circles. So there are $(n+k)!$ combinations. But we know that we can’t change the position of the circles and also the sticks are identical. So we have to divide $(n+k)!$ by $k!n!$ where $k!$ is the number of permutations of the sticks that we counted but shouldn’t have because they’re identical and where $n!$ is the number of permutations that the circles would have if we were able to change the position of the circles. So the answer would be: P(n+k;n,k)