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Some days ago Ilse Fischer has shown me a simple bijection. Associate with a tiling of an $n -$board with $k$ dominoes , $\ell$ black squares and $n - 2k - \ell$ white squares the word ${c_1}{c_2} \cdots {c_{n - k}}$ in the letters $w,b,d,$ where $d$ occurs $k$ times, $b$ occurs $\ell$ times and $w$ occurs $n - 2k - \ell$ times. Let $W({c_1}{c_2} \cdots {c_{n - k}})$ be the weight of the tiling.

First reverse in ${c_1}{c_2} \cdots {c_{n - k}}$ the order of the letters $b,d$ and obtain a word ${C_1}{C_2} \cdots {C_{n - k}}.$

Let e.g. $(n,k,\ell ) = (12,3,2)$ and ${c_1}{c_2} \cdots {c_9} = wbdwwdbwd.$ Then ${C_1}{C_2} \cdots {C_9} = wdbwwddwb.$

Then replace in ${C_1}{C_2} \cdots {C_{n - k}}$ all $b$ by $w.$ This gives a word $A$ with $k$ letters $d$ and $n - 2k$ letters $w.$ In our example we get $A = wdwwwddww.$

Then delete in ${C_1}{C_2} \cdots {C_{n - k}}$ all letters $d$ and get a word $B$ with $n - 2k$ letters $w,b.$ In our example $B = wbwwwb.$

Then $W({c_1}{c_2} \cdots {c_{n - k}}) = {q^{k\ell }}W(A)W(B).$

In our example we have $W({c_1}{c_2} \cdots {c_9}) = W(wbdwwdbwd) = {q^{2 + 3 + 7 + 9 + 11}} = {q^{32}},$ $W(A) = W(wdwwwddww) = {q^{2 + 7 + 9}} = {q^{18}},$ $W(B) = W(wbwwwb) = {q^{2 + 6}} = {q^8}.$

If $u(n,k,\ell )$ denotes the weighted enumeration of all tilings this implies$u(n,k,\ell ) = {q^{k\ell }}u(n,k,0)u(n - 2k,0,\ell ).$

1

Lucky chance or combinatorial cause?

Consider an $n \times 1 -$rectangle where the $n$ squares are numbered $1$ to $n$. Cover this rectangle with white squares, black squares, and dominoes. To each covering of the rectangle associate the following weight: Each white square has weight 1, each black square at position $i$ and each domino at position ${(i,i + 1)}$ have weight $q^i.$ As usual the weight of a covering is the product of its components and the weight of a set of coverings is the sum of their weights. Then it is easy to verify that the weight $u(n,k)$ of all coverings with precisely $k$ dominoes is the product $u(n,k)=a(n,k)b(n,k)$ with $a(n,k)= {q^{k^2}}{n-k\brack k}$ and $b(n,k) = (1 + {q^{k + 1}})(1 + {q^{k + 2}}) \cdots (1 + {q^{n - k}}).$

Here ${n\brack k}=[1][2]\dots[n]/(([1]\dots[k])([1]\dots[n-k]))$ with $[n]=(1-q^n)/(1-q)$ denotes a $q-$binomial coefficient.

It is often claimed that there are no accidents in mathematics. Therefore my question is: Is there a simple combinatorial reason for the fact that $u(n,k)$ is the product of two terms with simple combinatorial interpretations or is it an accident after all?

Since this seems to be rather elementary I have posted this question in http://math.stackexchange.com/questions/142158/lucky-chance-or-combinatorial-cause but did not get an answer.