TITLE: Summation with combinations
QUESTION [11 upvotes]: Prove that $n$ divides $$\sum_{d \mid \gcd(n,k)} \mu(d) \binom{n/d}{k/d}$$ for every natural number $n$ and for every $k$ where $1 \leq k \leq n.$ Note: $\mu(n)$ denotes the Möbius function.

I have tried numerous values for this summation and the result seems to hold true. For example, if $n = 20, k = 12$
$$\sum_{d \mid 4} \mu(d) \binom{20/d}{12/d} = \mu(1) \binom{20}{12}+\mu(2) \binom{10}{6}+\mu(4)\binom{5}{3} = \binom{20}{12}-\binom{10}{6}=125760,$$ which is divisible by $20$. Similarly if we tried it for any $k$ with $1 \leq k\leq 20$, we would have $20$ divide the expression.
How do I prove this result in the general case? That is, given any positive integer $n$, for all $k$ with $1 \leq k \leq n$ $$n \mid \sum_{d \mid \gcd(n,k)} \mu(d) \binom{n/d}{k/d}.$$

REPLY [5 votes]: I  would like  to present  the algebraic  aspects of  this  problem to
facilitate  understanding.   Suppose  we  have $r$  types  of  objects
(e.g. colors) with $k_1 + k_2 +  \cdots + k_r = n$ objects total where
$k_q$ gives  the number of objects of  color $q$ and we  ask about the
number of  necklaces we  can form with  these (rotational  symmetry as
opposed to dihedral symmetry).
Applying  the Polya  Enumeration Theorem  (PET) we  have  the cycle
index of the cyclic group
$$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}.$$
PET now yields for the  generating function of necklaces using at most
$r$ colors
$$q_n = \frac{1}{n}\sum_{d|n} 
\varphi(d) (A_1^d + A_2^d +\cdots+A_r^d)^{n/d}.$$
We  now  introduce  the  concept  of primitive  necklaces  $p_n$  i.e.
necklaces   on  at  most   $r$  colors   not  having   any  rotational
symmetry. Observe that an ordinary necklace is formed by concatenating
$d$ copies of a primitive necklace of size $n/d.$ (In fact it does not
matter  where we  open  the primitive  necklace ($n/d$  possibilities)
because when  we arrange the copies  of the opened  necklace we always
get  the same  necklace regardless  of where  we opened  the primitive
necklace.)
We   will   use  a   variety   of   Moebius   inversion  which   is
Inclusion-Exclusion on the divisor poset  in order to compute $p_n$ (a
generating function) and extract the desired coefficient. The possible
symmetries  that can  occur  correspond  to the  divisors  $f$ of  $n$
($f|n$).

Using  the variable $f$  we obtain  as explained  a segment  of length
$n/f$  being repeated  $f$ times,  copies  being placed  next to  each
other, thus creating $n/f$ cycles  of length $f.$.  These segments are
themselves  necklaces of  length $n/f.$  This means  that  the maximal
symmetry (smallest  size of  the constituent cycles)  is a  divisor of
$n/f$ because the segment could  itself be a concatenation of repeated
segments.  Ordering these in a poset by division yields an upside-down
instance  of the  divisor  poset  of $n$.   Note  that the  generating
function for the contribution from $f$ is not
$$\frac{1}{n/f}\sum_{d|n/f} 
\varphi(d) (A_1^d + A_2^d + \cdots + A_r^d)^{n/f/d}.$$
but rather
$$\frac{1}{n/f}\sum_{d|n/f} 
\varphi(d) (A_1^{df} + A_2^{df} + \cdots + A_r^{df})^{n/f/d}.$$
which represents the $f$ copies of the source segment.
We thus obtain by Inclusion-Exclusion
$$\sum_{f|n} \mu(f)
\frac{f}{n}\sum_{d|n/f} 
\varphi(d) (A_1^{df} + A_2^{df} + \cdots + A_r^{df})^{n/f/d}.$$
We put $fd=k$ so that $d=k/f$ to get
$$\sum_{f|n} \mu(f)
\frac{f}{n}\sum_{k/f|n/f} 
\varphi(k/f) (A_1^{k} + A_2^{k} + \cdots + A_r^{k})^{n/k}
\\ = \frac{1}{n}
\sum_{f|n} f\mu(f)
\sum_{k|n \wedge f|k} 
\varphi(k/f) (A_1^{k} + A_2^{k} + \cdots + A_r^{k})^{n/k}
\\ = \frac{1}{n}
\sum_{k|n} (A_1^{k} + A_2^{k} + \cdots + A_r^{k})^{n/k}
\sum_{f|k} f\mu(f) \varphi(k/f).$$
There are several ways to simplify the term
$$\sum_{f|k} f\mu(f) \varphi(k/f).$$
E.g. note that if
$$L_1(s) = \sum_{n\ge 1} \frac{n\mu(n)}{n^s}
= \prod_p \left(1-\frac{p}{p^s}\right) = \frac{1}{\zeta(s-1)}$$
and 
$$L_2(s) = \sum_{n\ge 1} \frac{\varphi(n)}{n^s} = 
\frac{\zeta(s-1)}{\zeta(s)}
\quad\text{because}\quad
\sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} \varphi(d)
= \zeta(s-1)$$
then
$$L_1(s) L_2(s) = \frac{1}{\zeta(s)}
\quad\text{and hence}\quad
\sum_{f|k} f\mu(f) \varphi(k/f) = \mu(k).$$ 
Substitute this into the formula to obtain
$$\frac{1}{n}
\sum_{k|n} \mu(k) (A_1^{k} + A_2^{k} +\cdots + A_r^{k})^{n/k}.$$
We seek
$$[A_1^{k_1} A_2^{k_2} \cdots A_r^{k_r}] \frac{1}{n}
\sum_{k|n} \mu(k) (A_1^{k} + A_2^{k} +\cdots + A_r^{k})^{n/k}.$$
Now observe that  the term in the variables  only produces powers that
are multiples of $k$ so we get the condition that
$$k|\gcd(n,  k_1, k_2,  \ldots k_r)$$
(we see that this produces a divisor of $n$) in which case we obtain a
contribution of (using $d$ for $k$ for readability)
$${n/d\choose k_1/d, k_2/d,\ldots k_r/d}$$
for an end result of
$$\frac{1}{n}\sum_{d|\gcd(k_1, k_2, \ldots k_r)} 
\mu(d) {n/d\choose k_1/d, k_2/d,\ldots k_r/d}.$$
We now  conclude by inspection  that the sum  is indeed a  multiple of
$n.$
A similar problem appeared at this
MSE link.