We denote the Littlewood-Richardson coefficient by $c_{\lambda, \mu}^{\nu}$ where $\mu$ is a partition of $m$ and $\nu$ is a partition of $n$ and $\lambda$ is a partition of $m+n$. We know $c_{\lambda, \mu}^{\nu}=\langle s_{\lambda}, s_{\eta}s_{\mu}\rangle$ where $s_{\lambda}$ is the Schur function corresponding to the partition $\lambda$. My question is about the determination of the numbers $c_{\lambda, \mu}^{(2)}$ and $c_{\lambda,\mu}^{(1,1)}$ where $\lambda$ is a partition of $n$ and $\mu$ is a partition of $n-2$. I get the following answer.
$$c_{\lambda,\mu}^{(2)}=\begin{cases} 1 & \text{when $\lambda$ can be obtained from $\mu$ by adding two boxes to the rows of the Young diagram of $\mu$} \\ 0 & \text{otherwise}.\end{cases}$$
Similarly, $$c_{\lambda,\mu}^{(1,1)}=\begin{cases} 1 & \text{when $\lambda$ can be obtained from $\mu$ by adding two boxes, but not in the same row, to the Young diagram of $\mu$} \\ 0 & \text{otherwise}.\end{cases}$$
Let us take an example to demonstrate this. We assume $\lambda=(4,2,2)\vdash 8$. We remove two boxes from the Young diagram of $\lambda$ such that it remains a Young diagram. The partitions of six that we get after doing that are $\{(2,2,2),(4,2),(4,1,1), (3,2,1)\}.$ Clearly these are all partitions of six such that if we add add two boxes to their Young diagram we get back $\lambda$. Thus, $$c_{(4,2,2),\mu}^{(2)}=\begin{cases} 1 & \text{when $\mu\in\{(2,2,2),(4,2),(4,1,1), (3,2,1)$\}} \\ 0 & \text{otherwise}.\end{cases}$$ By similar consideration, we get $$c_{(4,2,2),\mu}^{(1,1)}=\begin{cases} 1 & \text{when $\mu\in\{(4,1,1), (3,2,1)$\}} \\ 0 & \text{otherwise}.\end{cases}$$
Now the problem is that I found the following in a certain reference without any explanation.
$$c_{\lambda,\mu}^{(2)}=\begin{cases} 1 & \text{when $\lambda$ can be obtained from $\mu$ by adding two boxes to the columns, but not in the same column, in the Young diagram of $\mu$} \\ 0 & \text{otherwise}.\end{cases}$$
Considering this description we see that,
$$c_{(4,2,2),\mu}^{(2)}=\begin{cases} 1 & \text{when $\mu\in\{(2,2,2),(4,2),(3,2,1)$\}} \\ 0 & \text{otherwise}.\end{cases}$$
I don't understand where I am going wrong. The proof of my deduction is as follows:
We have, $$s_{\mu}s_{(1,1)}=\sum_{\lambda\vdash n}c_{\lambda,\mu}^{(1,1)}s_{\lambda}.$$ The coefficient $c_{\lambda,\mu}^{(1,1)}$ is the coefficient of $x^{\lambda+\delta}$ in $a_{\delta}s_{\mu}s_{(1,1)}$. But since $s_{\mu}=\frac{a_{\mu+\delta}}{a_{\delta}}$, we have to find the coefficient of $x^{\lambda+\delta}$ in $a_{\mu+\delta}s_{(1,1)}$. Note that $\delta=(n-1,\cdots,1,0)$ and $a_{\delta}$ is the Van-der-Monde determinant corresponding $\delta$.
Note that $s_{(1,1)}$ is just the elementary symmetric function $e_{2}=m_{(1,1)}$ where $m_{\eta}$ is the monomial symmetric function corresponding to a partition $\eta$. Thus we have to find the coefficient of $x^{\lambda+\delta}$ in $a_{\mu+\delta}e_{2}$. Since $a_{\mu +\delta}$ is a alternating polynomial, thus it's coefficient is determined by $x^{\mu+\delta}$ which is one. Thus the coefficient of $x_{\lambda+\delta}$ is $a_{\mu+\delta}e_2$ is one when $\lambda$ can be obtained from $\mu$ by adding one box each to two different rows of the Young diagram of $\mu$, zero otherwise.
The proof of $c_{\lambda,\mu}^{(2)}$ goes similarly by observing that $s_{(2)}=h_2$ where $h_2$ is the complete symmetric function corresponding to 2, that is, $h_2=m_2+m_{(1,1)}$. Note that because of the presence of $m_2$ we now get rid of the condition that two boxes are to be added in different rows.
Note that the above proof is based on the proof of Pieri's rule. Thank you in advance for any kind of assistance.