Counting Subsets of a Set: Combinations

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Counting Subsets of a Set Combinations

• Lecture 28
• Section 6.4
• Thu, Mar 30, 2006

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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.

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Counting r-Combinations

• Theorem The number of r-combinations of a set
  of n elements is
• 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.

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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.

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Another Useful Fact

• The TI-83 will calculate C(n, r).
• Enter n.
• Select MATH gt PRB gt nCr.
• Enter r.
• Press ENTER.
• The value of C(n, r) appears.

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Counting r-Combinations

• Proof (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.

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Counting r-Combinations

• General 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)!/(0!(k 1)!) 1.
• If r k 1, then there is only one
  k-combination, the entire set, and
• (k 1)!/((k 1)!0!) 1.

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Counting r-Combinations

• So let r be any number between 0 and k 1 (0 lt r
  lt k 1).
• Select an arbitrary element a from the set.
• 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.

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Counting r-Combinations

• r-combinations for which a is not a member
• The r elements come from the remaining k
  elements.
• Therefore, by the inductive hypothesis, there are
  k!/(r!(k r)!) such sets.
• r-combinations for which a is a member
• 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.

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Counting r-Combinations

• Therefore, the number of r-combinations of k 1
  elements is
• A little algebra shows that this equals
• Thus, the statement is true when n k 1, etc.

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Example Counting r-Combinations

• In Math 121, I used to collect 48 daily homework
  assignments.
• Some assignments count more than others.
• I drop the lowest 4 homework grades.
• Which should be dropped 0 out of 30 or 15 out
  of 40?
• I drop the 4 grades that hurt the students
  average the most.

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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.

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Example Counting r-Combinations

• What if I dropped 6 grades?
• There would be C(48, 6) 12,271,512 possible
  choices.
• Over 60 times as many.
• This would require about 60 secs 1 min.

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Example Counting r-Combinations

• What if I dropped 12 grades?
• There would be C(48, 12) over 69 billion
  choices!
• More than 350,000 as many!
• This would require almost 350,000 sec over 4
  days.

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Example Counting r-Combinations

• What if I dropped 44 grades!!!???
• This must involve unimaginably many
  possibilities!
• How many would it be?

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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Lotto South

• Match 0 of 6 numbers
• Player chooses 6 losing numbers.
• Number of ways is C(43, 6) 2760681.
• Probability is 0.4360.

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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.

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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.

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Example

• Theorem (The Vandermonde convolution) For all
  integers n ? 0 and for all integers r with 0 ? r
  ? n,
• Proof See p. 362, Sec. 6.6, Ex. 18.

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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?

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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

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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?

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Another Lottery

• What is the average prize value if matching 2
  numbers wins another lottery ticket?

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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!).

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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.

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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.

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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!).

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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.

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Example

• How many permutations are there of the letters in
  the word MISSISSIPPI?
• 11!/(4!4!2!1!) 34650.
• How many for VIRGINIA?
• How many for INDIVISIBILITY?