> Suppose there are $r$ things to be arranged, allowing repetitions. Let further $p_1,p_2,\dots p_r$ be the integers such that the first object occurs exactly $p_1$ times, the second occurs exactly $p_2$ times, etc. Then the total number of permutations of these $r$ objects to the above condition is $$\frac{(p_1+p_2+\dots+p_r)!}{p_1!p_2!\dots p_r!}$$
It seems $p_1+p_2+\dots+p_r\neq r$. So, I think the above formula doesn't count cases in which only one object of one kind is present and nothing else. Could anyone clarify whether I understood this properly? I understood the formula $\frac{n!}{p!q!}$ to determine the number of ways to arrange $n$ objects which contains two identical types of counts $p$ and $q$ each, where $p+q=n$
Or if possible, could anyone explain what the above formula means?
@GuruVishnu for googling purposes, that ratio is what's known as a multinomial coefficient, by analogy with the binomial coefficient which you also reference
as a starting point, consider the case of $r=3$ and note the following:
that is, you can interpret the left-hand term in the following way: "the number of ways to choose $p_1$ out of $p_1+p_2+p_3$ objects" times "the number of ways to choose $p_2$ out of $p_2+p_3$ objects"
which, if you think it it through, is exactly how you'd count the number of ways to take $p_1+p_2+p_3$ objects and divide them into boxes containing $p_1,p_2,p_3$ objects respectively
so that justifies the case of $r=3$. it's not too hard to turn that into a proof by induction if one puts in the work
