Computing the Möbius function of a poset 
Consider the multiset $ X = \{n_1\cdot a_1, ... ,n_k\cdot a_k \}$ of $k$ distinct elements with positive repetitions $n_1, n_2 ... n_k$. Define a partial order on $X$ with the following rule: if $A = \{p_1\cdot a_1, ... , p_k\cdot a_k\}$ and $B=\{q_1\cdot a_1, ... ,q_k\cdot a_k\}$ are combinations of $X$ then $A \leq B$ provided that $p_i \leq q_i$ for $ i =1,2,3 ...., k$. Prove that this is a partial order on $X$ and compute its Möbius function. 

I have showed that $(X,\le)$ is a poset, but I cannot figure out how to compute its Möbius function. Any help is appreciated. Also, if anyone can direct me to a good (easy enough for self-study) source where I can find more about the Möbius function and its uses, that would be great. Thanks!!
 A: The partial order isn’t actually defined on $X$: it’s defined on the family $\mathscr{X}$ of submultisets of $X$.
HINT: The Möbius function $\mu$ of this poset can be defined recursively as follows: $\mu(A,A)=1$ for each $A\in\mathscr{X}$, and for $A,B\in\mathscr{X}$ we have
$$\mu(A,B)=-\sum_{A\le Y<B}\mu(A,Y)\;.$$
Suppose that $A=\{p_1\cdot a_1,\ldots,p_k\cdot a_k\}\in\mathscr{X}$. Fix $\ell\in\{1,\ldots,k\}$, and assume that $p_\ell<n_\ell$. Let $B=\{q_1\cdot a_1,\ldots,q_k\cdot a_k\}$, where $q_j=p_j$ if $j\ne\ell$, and $q_\ell=p_\ell+1$; $B\in\mathscr{X}$, and $A<B$. However, no element of $\mathscr{X}$ lies strictly between $A$ and $B$ in the partial order: in the language of partial orders, $B$ covers $A$. Thus,
$$\mu(A,B)=-\sum_{A\le Y<B}\mu(A,Y)=-\mu(A,A)=-1\;,$$
since the only $Y\in\mathscr{X}$ satisfying the inequality $A\le Y<B$ is $Y=A$.
What happens if $p_\ell+2\le n_\ell$, and we let $C=\{r_1\cdot a_1,\ldots,r_k\cdot a_k\}$, where $r_j=p_j$ if $j\ne\ell$, and $r_\ell=p_\ell+2$? Then the only multisets in the interval $[A,C]=\{Y\in\mathscr{X}:A\le Y\le C\}$ are $A,B$, and $C$, so
$$\mu(A,C)=-\sum_{A\le Y<C}\mu(A,Y)=-\big(\mu(A,A)+\mu(A,B)\big)=-\big(1+(-1)\big)=0\;.$$
Alternatively, what happens if we choose distinct $\ell,m\in\{1,\ldots,k\}$ such that $p_\ell<n_\ell$ and $p_m<n_m$ and let $D=\{r_1\cdot a_1,\ldots,r_k\cdot a_k\}$, where $r_j=p_j$ if $\ell\ne j\ne m$, $r_\ell=p_\ell+1$, and $r_m=p_m+1$. Then the interval $[A,D]$ has four elements: $A,B,D$, and a multiset $B'$ similar to $B$, but with the count of $a_m$ increased by $1$ instead of the count of $a_\ell$. Then
$$\begin{align*}
\mu(A,D)&=-\sum_{A\le Y<D}\mu(A,Y)\\
&=-\big(\mu(A,A)+\mu(A,B)+\mu(A,B')\big)\\
&=-\big(1+(-1)+(-1)\big)\\
&=1\;.
\end{align*}$$
Now let $A=\{p_1\cdot a_1,\ldots,p_k\cdot a_k\}$ and $B=\{q_1\cdot a_1,\ldots,q_k\cdot a_k\}$ be any elements of $\mathscr{X}$. If $A\not\le B$, $\mu(A,B)=0$. For the more interesting case in which $A\le B$, the calculations that I made above illustrate the basics; try to use them as models to express $\mu(A,B)$ in terms of the numbers $q_j-p_j$ for $j=1,\ldots,k$.
I suggest that you first investigate the case $k=1$; in this case it it’s quite easy to figure out what happens and to prove it. Then experiment with the case $k=2$; once you reach the right conjecture there, it’s a pretty short step to the case of arbitrary $k$. I’ve left the correct result, without proof, in the spoiler-protected block below.

 If $q_j<p_j$ or $q_j-p_j>1$ for some $j$, then $\mu(A,B)=0$. Otherwise, let $K=\{j:q_j-p_j=1\}$; then $\mu(A,B)=(-1)^{|K|}$.

