Counting by liaoqinmei

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									Counting Subsets of a
Set: Combinations

Lecture 31
Section 6.4
Wed, Mar 21, 2007
     r-Combinations

 An r-combination of a set of n elements is a
  subset of r of the n elements.
 The order of the elements does not matter.
 The 3-combinations of the set {a, b, c, d, e}
  are
     {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d},
     {a, c, e}, {a, d, e}, {b, c, d}, {b, c, e},
                {b, d, e}, {c, d, e}.
        Counting r-Combinations

   Theorem: The number of r-combinations of
    a set of n elements is
                                 n!
                 C (n, r ) 
                             r!(n  r )!
   Examples:
     C(4, 2) = (4  3)/(2  1) = 6.
     C(10, 3) = (10  9  8)/(3  2  1) = 120.
     C(1000, 2) = (1000  999)/(2  1) = 499500.
        Some Useful Facts

 C(n, 0) = 1 for all n  0.
 C(n, 1) = n for all n  1.
 Notice that C(n, r) = C(n, n – r).
 For example,
       C(100, 99) = C(100, 1) = 100/1 = 100.
   Therefore,
     C(n, n) = 1 for all n  0.
     C(n, n – 1) = n for all n  1.
        Another Useful Fact

   The TI-83 will calculate C(n, r).
     Enter n.
     Select MATH > PRB > nCr.
     Enter r.
     Press ENTER.
     The value of C(n, r) appears.
        Counting r-Combinations

 Proof of the theorem (by induction on n).
 Base case:
     Let n = 0. Then r = 0 and there is only one
      0-combination, the null set.
     Also, 0!/(0!0!) = 1.
     So the statement is true when n = 0.
        Counting r-Combinations

   Inductive case:
     Suppose that the statement is true when n =
      k, for some integer k  0.
     Consider a set of k + 1 elements.
     If r = 0, then there is only one 0-combination,
      the null set, and
                       k  1!  1.
                      0!k  1!
    Counting r-Combinations

   If r = k + 1, then there is only one k-
    combination, the entire set, and
                     k  1!  1.
                    k  1!0!
 So let r be any number between 0 and k + 1
  (0 < r < k + 1).
 Select an arbitrary element a from the set.
    Counting r-Combinations

 For each r-combination of the k + 1
  elements, a is either a member or not a
  member.
 We will count the r-combinations for which
  a is a member and then count the r-
  combinations for which a is not a member.
    Counting r-Combinations

   Case 1: a is not a member of the
    combination:
     • The r elements come from the remaining k
       elements.
     • By the inductive hypothesis, there are
                           k!
                       r!k  r !
      such sets.
    Counting r-Combinations

   Case 2: a is a member of the combination:
     • The other r – 1 elements in the subset come from
       the k remaining elements in the set.
     • By the inductive hypothesis, there are
                              k!
                    r  1!k  r  1!
      such sets.
    Counting r-Combinations

   Therefore, the number of r-combinations of k
    + 1 elements is
                  n!                 n!
                         
              r!(n  r )! (r  1)! (n  (r  1))!
   A “little algebra” shows that this equals
                           (n  1)!
                       r!(( n  1)  r )!
    Counting r-Combinations

 Therefore, the statement is true when n = k +
  1.
 Thus, the statement is true for all n  1.
     Example: Counting r-
     Combinations
 Recently I needed to find the distribution of
  averages of 10 numbers selected at
  random from a set of 19 numbers.
 I wrote a C++ program to use brute force to
  calculate the distribution.
 It is much easier to write the program if the
  sampling is done with replacement.
     Example: Counting r-
     Combinations
 Sampling with replacement, there are 1910
  possible samples.
           1910 = 6131066257801.
 The program took 21.2 seconds to
  compute the distribution using 7 instead of
  10 numbers.
 How long would it take using 10 numbers?
     Example: Counting r-
     Combinations
 How many possibilities are there if we
  sample without replacement?
 How long would it take to calculate the
  distribution?
     Example: Counting r-
     Combinations
 How can that be determined?
 Can a computer program make the
  determination by brute force (exhaustive
  checking) within a reasonable amount of
  time?
 There are C(48, 4) = 194,580 possible
  choices.
 A computer can do the math really fast, in
  say one second.
    Lotto South

 In Lotto South, a player chooses 6
  numbers from 1 to 49.
 Then the state chooses at random 6
  numbers from 1 to 49.
 The player wins according to how many of
  his numbers match the ones the state
  chooses.
 See the Lotto South web page.
        Lotto South

 There are C(49, 6) = 13,983,816 possible
  choices.
 Match all 6 numbers
     There is only 1 winning combination.
     Probability of winning is
           1/13983816 = 0.00000007151.
        Lotto South

   Match 5 of 6 numbers
     There are 6 winning numbers and 43 losing
      numbers.
     Player chooses 5 winning numbers and 1
      losing numbers.
     Number of ways is C(6, 5)  C(43, 1) = 258.
     Probability is 0.00001845.
        Lotto South

   Match 4 of 6 numbers
     Player chooses 4 winning numbers and 2
      losing numbers.
     Number of ways is C(6, 4)  C(43, 2) =
      13545.
     Probability is 0.0009686.
        Lotto South

   Match 3 of 6 numbers
     Player chooses 3 winning numbers and 3
      losing numbers.
     Number of ways is C(6, 3)  C(43, 3) =
      246820.
     Probability is 0.01765.
        Lotto South

   Match 2 of 6 numbers
     Player chooses 2 winning numbers and 4
      losing numbers.
     Number of ways is C(6, 2)  C(43, 4) =
      1851150.
     Probability is 0.1324.
        Lotto South

   Match 1 of 6 numbers
     Player chooses 1 winning numbers and 5
      losing numbers.
     Number of ways is C(6, 1)  C(43, 5) =
      3011652.
     Probability is 0.4130.
        Lotto South

   Match 0 of 6 numbers
     Player chooses 6 losing numbers.
     Number of ways is C(43, 6) = 2760681.
     Probability is 0.4360.
        Lotto South

 Note also that the sum of these integers is
  13983816.
 Note also that the lottery pays out a prize
  only if the player matches 3 or more
  numbers.
     Match 3 – win $5.
     Match 4 – win $75.
     Match 5 – win $1000.
     Match 6 – win millions.
     Lotto South

 Given that a lottery player wins a prize,
  what is the probability that he won the $5
  prize?
 P(he won $5, given that he won)

      = P(match 3)/P(match 3, 4, 5, or 6)
      = 0.01765/0.01864
      = 0.9469.
       Example

   Theorem (The Vandermonde convolution):
    For all integers n  0 and for all integers r
    with 0  r  n,
                r
                     r  n  r   n 
                 k  r  k    r 
                     
               k 0  
                                   
                                   

   Proof: See p. 362, Sec. 6.6, Ex. 18.
     Another Lottery

 In the previous lottery, the probability of
  winning a cash prize is 0.018637545.
 Suppose that the prize for matching 2
  numbers is… another lottery ticket!
 Then what is the probability of winning a
  cash prize?
        Lotto South

 What is the average prize value of a ticket?
 Multiply each prize value by its probability
  and then add up the products:
     $10,000,000  0.00000007151 = 0.7151
     $1000  0.00001845 = 0.0185
     $75  0.0009686 = 0.0726
     $5  0.01765 = 0.0883
     $0  0.9814 = 0.0000
    Lotto South

 The total is $0.8945, or 89.45 cents
  (assuming that the big prize is ten million
  dollars).
 A ticket costs $1.00.

 How large must the grand prize be to make
  the average value of a ticket more than
  $1.00?
      Another Lottery

   What is the average prize value if matching
    2 numbers wins another lottery ticket?
      Permutations of Sets with
      Repeated Elements
   Theorem: Suppose a set contains n1
    indistinguishable elements of one type, n2
    indistinguishable elements of another type,
    and so on, through k types, where
               n1 + n2 + … + nk = n.
    Then the number of (distinguishable)
    permutations of the n elements is
                  n!/(n1!n2!…nk!).
        Proof of Theorem

   Proof:
     Rather than consider permutations per se,
      consider the choices of where to put the
      different types of element.
     There are C(n, n1) choices of where to
      place the elements of the first type.
        Proof of Theorem

   Proof:
     Then there are C(n – n1, n2) choices of
      where to place the elements of the second
      type.
     Then there are C(n – n1 – n2, n3) choices of
      where to place the elements of the third
      type.
     And so on.
       Proof, continued

   Therefore, the total number of choices, and
    hence permutations, is
     C(n, n1)  C(n – n1, n2)  C(n – n1 – n2, n3)
         … C(n – n1 – n2 – … – nk – 1, nk)
       = …(some algebra)…
       = n!/(n1!n2!…nk!).
     Example

 How many different numbers can be
  formed by permuting the digits of the
  number 444556?
 6!/(3!2!1!) = 720/(6  2  1) = 60.
      Example

   How many permutations are there of the
    letters in the word MISSISSIPPI?
                   11!
                             34650
                4!4!2!1!

 How many for VIRGINIA?
 How many for INDIVISIBILITY?
       Poker Hands

   Two of a kind.
   Two pairs.
   Three of a kind.
   Straight.
   Flush.
   Full house.
   Four of a kind.
   Straight flush.
   Royal flush.

								
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