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:-
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:-
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.