that $\text{comaj}(P)=\ell$. Since $$A(2,n)=\{(i,i+2),(p,p+1): 0\leq i\leq n_1-1,\ n_1<p\leq n-1\}$$ we see that any admissible path has no two adjacent down–steps and the last $n_1$ steps are up–steps. Hence $P$ is uniquely determined by the $x$–coordinates of all its peaks and the number of down–steps is equal to the number of peaks (= $k$). Assume that $i_1,\dots,i_{k}$ are these $x$–coordinates, then we must have $i_{p+1}\geq i_p+2$ for all $1\leq p<k$ and $$1\leq i_1<i_2<\cdots<i_{k}<2k+n_1,\ \ \sum_{p=1}^{k} i_p=(n+2k)k-\ell.$$ Now substituting $i_p'=i_p-2p+1$ we get a partition $$0\leq i'_1\leq i'_2\leq \cdots\leq i'_{k}\leq n_1, \ \ \sum_{p=1}^{k} i'_p=k(k+n)-\ell,$$ which obviously gives a bijective correspondence $$\{P\in \mathcal{D}_{2,n}(n+2k) : \text{comaj}(P)=\ell\}\cong \rho^{k}_{n_1}(k(k+n)-\ell).$$

 {#section-9}

Our next corollary answers the following question. What is the set of all powers $q$ which appear with nonzero coefficients in the polynomial $\mathcal{V}^{1\rightarrow m}_{s,n}(q)$? We first determine the degree of $\mathcal{V}^{1\rightarrow m}_{s,n}(q)$ as we did in Lemma \[lpart\] for $m=2$. By we can assume that $s=n+2k$ for some $k\in {\mathbb Z}_+$. The path $$\label{4}P=\underbrace{1010\cdots 1010}_{2k}\underbrace{111\cdots 11}_{n}$$ with exactly $k$ up–down steps at the beginning and $n$ up–steps at the end is contained in $\mathcal{D}_n(s)$. Note that all peaks of $P$ are at height $1$ and $(0,1)\notin A(m,n)$, which implies that $P$ is admissible. Moreover, $$\label{2}\text{comaj}(P)=\sum_{i=1}^k(s-2i+1)=k(n+k).$$ 