I’m looking at partition numbers. OEIS A008284 says that the number of partitions of $n$ in which the greatest part is $k$, $1 \le k \le n$, is equal to the number of partitions of $n$ into $k$ positive parts ($1 \le k \le n$). (This also easily follows from Ferrers diagrams.)
However, I want to find out how many partitions of $n$ are if the greatest part is at most $k_1$ and the number of positive parts is $k_2$, where possibly $k_1 \ne k_2$. For instance, for $n = 8, k_1 = 4, k_2 = 3$, we have 431, 422, and 332. Is there anything known about this?
Actually … I’m not only interested in this number, but in the number of permutations of these partitions. Since 431 has six permutations and both 422 and 332 have three, the number I’m interested in in the above example would be 12.
I couldn’t find anything at all even about permutations in combination with permutation numbers, let alone my additional restrictions above. Maybe someone who’s a bit more into number theory can help?
The number of partitions of $n$ into $k_2$ positive parts, no part exceeding $k_1$, is the coefficient of $x^n$ in $(x+x^2+x^3+\cdots+x^r)^s$, where I have written $r$ for $k_1$, and $s$ for $k_2$. This is the coefficient of $x^t$ in $(1+x+x^2+\cdots+x^{r-1})^s$, where I have written $t$ for $n-s$. Now $$(1+x+x^2+\cdots+x^{r-1})^s=(1-x^r)^s(1-x)^{-s}=\sum_0^s(-1)^p{s\choose p}x^{rp}\sum_0^{\infty}{s-1+q\choose s-1}x^q$$ so the answer is the sum, over all pairs $p,q$ such that $rp+q=t$, of $(-1)^p{s\choose p}{s-1+q\choose s-1}$.
Try this out on your original example. $n=8$, $k_1=r=4$, $k_2=s=3$, $t=n-s=5$; the solutions of $4p+q=5$ are $p=0$, $q=5$ and $p=1$, $q=1$, so you get $$(-1)^0{3\choose0}{7\choose2}+(-1)^1{3\choose1}{3\choose2}=(1)(21)-(3)(3)=21-9=12$$
Note there are no trinomial, quadrinomial, whatever coefficients, just binomials.