auto-update(nvim): 2025-01-19 22:00:39

documents/by-course/pstat-120a/course-notes/main.typ

@@ -580,7 +580,7 @@ Now we can use the multiplication principle to count permutations.
   $ (n)_k = n dot (n - 1) ... (n - k + 1) = n! / (n-k)! $
 
   In particular, with $k=n$, each $n$-tuple is an ordering or _permutation_ of $A$. So the total number of permutations of a set of $n$ elements is $n!$.
-]
+]<permutation>
 
 #proof[
   We construct the $k$-tuples sequentially. For the first element, we choose
@@ -592,3 +592,106 @@ Now we can use the multiplication principle to count permutations.
   @tuplemultiplication the number of $k$-tuples being $n dot (n - 1) dot ...
   dot (n - k + 1) = (n)_k$.
 ]
+
+#example[
+  Consider a round table with 8 seats.
+
+  + In how many ways can we seat 8 guests around the table?
+  + In how many ways can we do this if we do not differentiate between seating arrangements that are rotations of each other?
+
+  For (1), we easily see that we're simply asking for permutations of an
+  8-tuple, so $8!$ is the answer.
+
+  For (2), we number each person and each seat from 1-8, then always place person 1 in seat 1, and count the permutations of the other 7 people in the other 7 seats. Then the answer is $7!$.
+
+  Alternatively, notice that each arrangement has 8 equivalent arrangements under rotation. So the answer is $8!/8 = 7!$.
+]
+
+== Counting from sets
+
+We turn our attention to sets, which unlike tuples are unordered collections.
+
+#fact[
+  Let $n,k in NN$ with $0 <= k <= n$. The numbers of distinct subsets of size $k$ that a set of size $n$ has is given by the *binomial coefficient*
+  $ vec(n,k) = n! / (k! (n-k)!) $
+]
+
+#proof[
+  Let $A$ be a set of size $n$. By @permutation, $n!/(n-k)!$ unique ordered
+  $k$-tuples can be constructed from elements of $A$. Each subset of $A$ of
+  size $k$ has exactly $k!$ different orderings, and hence appears exactly $k!$
+  times among the ordered $k$-tuples. Thus the number of subsets of size $k$ is
+  $n! / (k! (n-k)!)$.
+]
+
+#example[
+  In a class there are 12 boys and 14 girls. How many different teams of 7 pupils
+  with 3 boys and 4 girls can be create?
+
+  First let us compute how many subsets of size 3 we can choose from the 12 boys and how many subsets of size 4 we can choose from the 14 girls.
+
+  $
+    "boys" &= vec(12,3) \
+    "girls" &= vec(14,4)
+  $
+
+  Then let us consider the entire team as a 2-tuple of (boys, girls). Then
+  there are $vec(12,3)$ alternatives for the choice of boys, and $vec(14,4)$ alternatives for
+  the choice of girls, so by the multiplication principle, we have the total being
+
+  $ vec(12,3) vec(14,4) $
+]
+
+#example[
+  Color the numbers 1, 2 red, the numbers 3, 4 green, and the numbers 5, 6
+  yellow. How many different two-element subsets of $A$ are there that have two
+  different colors?
+
+  First choose 2 colors, $vec(3,2) = 3$. Then from each color, choose one. Altogether it's
+  $ vec(3,2) vec(2,1) vec(2,1) = 3 dot 2 dot 2 = 12 $
+]
+
+One way to view $vec(n,k)$ is as the number of ways of painting $n$ elements
+with two colors, red and yellow, with $k$ red and $n - k$ yellow elements. Let
+us generalize to more than two colors.
+
+#fact[
+  Let $n$ and $r$ be positive integers and $k_1, ..., k_r$ nonnegative integers
+  such that $k_1 + dots.c + k_r = n$. The number of ways of assigning labels
+  $1,2, ..., r$ to $n$ items so that for each $i = 1, 2, ..., r$, exactly $k_i$
+  items receive label $i$, is the *multinomial coefficient*
+
+  $ vec(n, (k_1, k_2, ..., k_r)) = vec(n!, k_1 ! k_2 ! dots.c k_r !) $
+]<multinomial-coefficient>
+
+#proof[
+  Order the $n$ integers in some manner, and assign labels like this: for the
+  first $k_1$ integers, assign the label 1, then for the next $k_2$ integers,
+  assign the label 2, and so on. The $i$th label will be assigned to all the
+  integers between positions $k_1 + dots.c + k_(i-1) + 1$ and $k_1 + dots.c +
+  k_i$.
+
+  Then notice that all possible orderings (permutations) of the integers gives
+  every possible way to label the integers. However, we overcount by some
+  amount. How much? The order of the integers with a given label don't matter,
+  so we need to deduplicate those.
+
+  Each set of labels is duplicated once for each way we can order all of the
+  elements with the same label. For label $i$, there are $k_i$ elements with
+  that label, so $k_i !$ ways to order those. By @tuplemultiplication, we know
+  that we can express the combined amount of ways each group of $k_1, ..., k_i$
+  numbers are labeled as $k_1 ! k_2 ! k_3 ! dots.c k_r !$.
+
+  So by @ktoone, we can account for the duplicates and the answer is
+  $ n! / (k_1 ! k_2 ! k_3 ! dots.c k_r !) $
+]
+
+#remark[
+  @multinomial-coefficient gives us a way to count how many ways there are to
+  fit $n$ distinguishable objects into $r$ distinguishable containers of
+  varying capacity.
+
+  To find the amount of ways to fit $n$ distinguishable objects into $k$
+  indistinguishable containers of equal capacity, use the "ball-and-urn"
+  technique.
+]