## Multisets

Example:
    A bag of Scrabble tiles contains 100 tiles: 10 A's, 2 B's, 2 C's, 5 D's
    and so on.

    When you start a game, you take 7 letters from the bag, and put them on a
    rack. How many possible hands can you get, if we say that the order of the
    tiles on the rack matters? How about if it doesn't?

A _multiset_ is a "set with multiplicity".

Notation: $\{2*a,3*b,1*c\}$

(Think of a "bag" with 2 $a$'s, 3 $b$'s and a $c$ in it.)

We also allow "infinite multiplicity", denoted $\{\infty*a\}$.

Multiplicities are also called "repetition numbers".

The _size_ of a multiset is the sum of the multiplicities (may be $\infty$).

An $r$-permutation of a multiset is an ordered list of $r$ elements from the
multiset;
e.g. the 2-permutations of $\{2*a,1*b\}$ are
    :	$aa, ab, ba$;
the 3-permutations of $\{\infty*a,2*b\}$ are
    :	$aaa, aab, aba, baa, abb, bab, bba$.

A _permutation_ of a multiset of size $n$ is an $n$-permutation.

Example: how many permutations are there of the unfortunate scrabble hand
    $\{4*U,1*J,2*K\}$?

Theorem:
    Let $S$ be a multiset with $k$ types with finite multiplicities $n_1,...,n_k$.
    Let $n = \Sigma_i n_i$ be the size of $S$.
    Then the number of permutations of $S$ is
    :	$n! / n_1!*n_2!*...*n_k!$.
Proof:
    Label the elements $1,...,n$. Each permutation of $\{1,...,n\}$ yields a
    permutation of $S$, and two yield the same permutation precisely when we can
    get one from another by permuting the labels on elements of the same type.
    So there are $n_1!*n_2!*...*n_k$! permutations of $\{1,...,n\}$ per permutation
    of $S$. We conclude by the division principle.
    
Example:
    We have 4 black rooks and 4 white rooks. How many ways are there of
    putting them on a chess board such that no two are attacking (/defending)
    each other? e.g.

\begin{verbatim}
    .......R  8
    ....r...  5
    ..R.....  3
    ......R.  7
    ...r....  4
    .R......  2
    .....r..  6
    r.......  1
\end{verbatim}
Solution:
    First, just choose the 8 squares for them to occupy.

    By listing off the filled columns row-by-row, a choice corresponds to a
    permutation of the columns, so there are 8!.

    For a given such choice, a choice of colours corresponds to a permutation
    of the multiset $\{4*r,4*R\}$.

    So the answer is
    :	$8! * (8! / 4!*4!) = 2822400$


An \defn{$r$-submultiset}, or \defn{$r$-combination}, of a multiset $S$ is a
multiset $S'$ of size $r$ such that for all $x$,
    the multiplicity of $x$ in $S'$ is at most the multiplicity of $x$ in $S$.

e.g. the 2-submultisets of $\{3*a,b\}$ are
    $\{2*a\}$, $\{1*a,1*b\}$.

\pagebreak
Theorem:
    The number of $r$-combinations of a multiset with $k$ types each with
    multiplicity at least $r$ is
    :	${r+k-1 \choose r}$
Example:
    If we have a bag containing red, green and blue marbles, with many of
    each, and we draw 5 marbles from the bag, how many possible results
    (numbers of each colour drawn) are there?
    Answer: ${5+3-1 \choose 5} = {7 \choose 5} = 7! / 2!5! = 21$
Proof:
    We can identify an $r$-submultiset with an arrangement of $k-1$ partitions
    interspersed among $r$ identical objects, by counting the numbers of objects
    between the partitions; e.g. with $k=6$ and $r=8$
    \begin{verbatim}
	oo|ooo||o|oo|
    \end{verbatim}
    corresponds to
    :	$\{ 2*a_1, 3*a_2, 0*a_3, 1*a_4, 2*a_5, 0*a_6 \}$.

    These arrangements correspond to choosing $r$ of the $r+k-1$ characters to
    be 'o's, so the number of such arrangements is ${r+k-1 \choose r}$.

