Can we generalize it?

I was working on one question:- if there are 1 red balls, 2 blue balls, 3 green balls, 4 yellow balls and 5 white balls placed on a table, what is the number of ways of selecting 5 out of 5. My process came out with 7 cases as follows:-

5 alike balls.
4 alike balls and 1 different ball.
3 alike balls and 2 different balls.
3 alike balls of one colour and 2 alike balls of another colour.
2 alike balls and 3 different balls.
2 alike balls of one colour, 2 alike balls of another colour and 1 different ball.
5 different balls.

I got interested and tried to generalize the number of such cases that would form if we have to select $n$ balls from 1 ball of colour-1, 2 balls of colour-2, 3 balls of colour-3, $\cdots n$ balls of colour $n$ .

It turns out that when $n = 1$ , the number of cases is $1$ .
When $n = 2$ , the number of cases is $2$ .
When $n = 3$ , the number of cases is $3$ .
When $n = 4$ , the number of cases is $5$ .
When $n = 5$ , the number of cases is $7$ . (as discussed above).
When $n = 6$ , the number of cases is $12$ .

Now, just in case that the sixth case becomes clear, here are the cases I found:-

6 alike.
5 alike and 1 different.
4 alike and 2 different.
4 alike and 2 alike.
3 alike and 3 different.
3 alike and 2 alike and 1 different.
3 alike and 3 alike.
2 alike and 4 different.
2 alike and 3 alike and 1 different.
2 alike and 2 alike and 2 different.
2 alike and 2 alike and 2 alike.
6 different.

Looking at the pattern, the first thing which came into my mind is the fibonacci sequence, but as you can observe, the pattern is broken at $n = 5$ . So, is there any nice method to calculate the number of cases that can be formed?

A detailed solution is appreciated.

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Can we generalize it?

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