inclusion exclusion principle, how many ways to order in 8 places 4 labels

Question

8 places 4 labels $\{1,2,3,4\}$

the original order 1,1,2,2,3,3,4,4 how many ways to order such that no label will have the original position.

solution:

we will use inclusion exclusion principle , we will find all ways such that they are in the original places
A1- we have $6!/(2!2!2!)$ and for the same logic we get : $A_i$ = $4\choose1$$6!/(2!2!2!)$

$A_(1\cap2)$= $4!/(2!2!)$ and by the same logic we get : $A_(i\cap j)$= $4\choose2$$4!/(2!2!)$, $i\neq j$

$A_(1\cap2\cap3)$=$2!/(2!)$ and by the same logic we get: $A_(i\cap j\cap k)$=$4\choose3$$2!/(2!)$ where i,j,K differ from one another
for 4 of them we have 1 option\

we conclude : $|U|-(A_i)+(A_(i\cap j))-(A_(i\cap j \cap k )) +...= $$8!/(2!2!2!2!)$-$4\choose1$$6!/(2!2!2!)$+$4\choose2$$4!/(2!2!)$-$4\choose3$$2!/(2!)$+1

is it Correct?

Answer 1

This is a special case of this problem.

You can compute the result by counting the number of placements of 8 rooks on the $8\times 8$ board with four $2\times 2$ squares on the diagonal removed and then dividing the result by $(2!)^4$.

First find the rook polynomial of the complement of this board, which consists of four independent $2\times 2$ squares, hence the polynomial is $$(1+4x+x^2)^4 = 16 x^8 + 128 x^7 + 416 x^6 + 704 x^5 + 664 x^4 + 352 x^3 + 104 x^2 + 16 x + 1$$

Now apply the inclusion-exclusion formula for the number of rooks on the original board $$\sum_{k} (-1)^k(8-k)!r_k,$$ where $r_k$ is the coefficient of $x^k$ in the above polynomial (it is fully explained in the answer I linked), to get $$1\cdot 8!-16\cdot 7!+104\cdot 6!-352\cdot5!+664\cdot 4!-704\cdot 3!+416\cdot 2!-128\cdot 1!+16\cdot 0! = 4752$$

The number of rook placements is $(2!)^4$-times bigger than the number you seek because the rooks are labeled by their row numbers (so, e.g. the placement in which the rooks in rows 1 and 2 are in the columns 3 and 4, respectively, is different from the placement where these rooks are in columns 4 and 3, but they both correspond to a single permutation ##11####).

So the final result is $4752/(2!)^4 = 297$.

Answer 2

No, it is not correct.

You seem to exclude only the cases when both similar labels stay on their places. One should however exclude also the cases when only one of two labels is on one of two places. And this is much more complicated.

Let count the cases where at least $k$ of $n$ labels stay on their places. The number of such combinations will be: $$ D(n,k)=\sum_{i=0}^k (-1)^i\binom ki 2^{k-i}\frac{(2n-k-i)!}{(2!)^{n-k}}, $$ where $i$ is the number of pairs with both labels being on their places, the factor $2^{k-i}$ counts the number of ways to choose the positions of the "persistent" labels in the pairs with only one such label, and $\frac{(2n-k-i)!}{(2!)^{n-k}}$ counts the number of ways to arrange all other labels. The factor $(-1)^i \binom ki$ performs here the inclusion-exclusion principle, taking into account the number of ways to choose $i$ pairs out of $k$ ones.

Finally the number of ways in question is: $$ R(n)=\sum_{k=0}^n(-1)^k\binom nk D(n,k). $$

Particularly for $n=4$ it gives $297$.