auto-update(nvim): 2025-01-19 23:14:43

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

@@ -692,6 +692,82 @@ us generalize to more than two colors.
   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"
+  indistinguishable containers of _any_ capacity, use the "ball-and-urn"
   technique.
 ]
+
+#example[
+  How many different ways can six people be divided into three pairs?
+
+  First we use the multimonial coefficient to count the amount of ways to assign specific labels to pairs of elements:
+  $ vec(6, (2,2,2)) $
+  But notice that the actual labels themselves are irrelevant. Our multimonial
+  coefficient counts how many ways there are to assign 3 distinguishable
+  labels, say Pair 1, Pair 2, Pair 3, to our 6 elements.
+
+  To make this more explicit, say we had a 3-tuple where the position encoded
+  the label, where position 1 corresponds to Pair 1, and so on. Then the values
+  are the actual pairs of people (numbered 1-6). For instance
+  $ ((1,2), (3,4), (5,6)) $
+  corresponds to assigning the label Pair 1 to (1,2), Pair 2 to (3,4) and Pair
+  3 to (5,6). What our multimonial coefficient is doing is it's counting this,
+  as well as any other orderings of this tuple. For instance
+  $ ((3,4), (1,2), (5,6)) $
+  is also counted. However since in our case the actual labels are irrelevant,
+  the two examples shown above should really be counted only once.
+
+  How many extra times is each case counted? It turns out that we can think of
+  our multimonial coefficient as permuting the labels across our pairs. So in
+  this case it's permuting all the ways we can order 3 labels, which is $3! =
+  6$. That means by @ktoone our answer is
+
+  $ vec(6, (2,2,2)) / 3! = 15 $
+]
+
+#example("Poker")[
+  How many poker hands are in the category _one pair_?
+
+  A one pair is a hand with two cards of the same rank and three cards with ranks
+  different from each other and the pair.
+
+  We can count in two ways: we count all the ordered hands, then divide by $5!$
+  to remove overcounting, or we can build the unordered hands directly.
+
+  When finding the ordered hands, the key is to figure out how we can encode
+  our information in a tuple of the form described in @tuplemultiplication, and
+  then use @tuplemultiplication to compute the solution.
+
+  In this case, the first element encodes the two slots in the hand of 5 our
+  pair occupies, the second element encodes the first card of the pair, the
+  third element encodes the second card of the pair, and the fourth, fifth, and
+  sixth elements represent the 3 cards that are not of the same rank.
+
+  Now it is clear that the number of alternatives in each position of the
+  6-tuple does not depend on any of the others, so @tuplemultiplication
+  applies. Then we can determine the amount of alternatives for each position
+  in the 6-tuple and multiply them to determine the total amount of ways the
+  6-tuple can be constructed, giving us the total amount of ways to construct
+  ordered poker hands with one pairs.
+
+  First we choose 2 slots out of 5 positions (in the hand) so there are
+  $vec(5,2)$ alternatives. Then we choose any of the 52 cards for our first
+  pair card, so there are 52 alternatives. Then we choose any card with the
+  same rank for the second card in the pair, where there are 3 possible
+  alternatives. Then we choose the third card which must not be the same rank
+  as the first two, where there are 48 alternatives. The fourth card must not
+  be the same rank as the others, so there are 44 alternatives. Likewise, the
+  final card has 40 alternatives.
+
+  So the final answer is, remembering to divide by $5!$ because we don't care
+  about order,
+  $ (vec(5,2) dot 52 dot 3 dot 48 dot 44 dot 40) / 5! $
+
+  Alternatively, we can find way to build an unordered hand with the
+  requirements. First we choose the rank of the pair, then we choose two suits
+  for that rank, then we choose the remaining 3 different ranks, and finally a
+  suit for each of the ranks. Then, noting that we will now omit constructing
+  the tuple and explicitly listing alternatives for brevity, we have
+  $ 13 dot vec(5,2) dot vec(12, 3) dot 4^3 $
+
+  Both approaches given the same answer.
+]