Counting Techniques: Combinations

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Counting TechniquesCombinations
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Combinations

• Typical situation How many 5-card hands
• can be made from a deck of 52 cards?
• Definition r-combination of a set of n elements
  is a subset of r of the n elements.
• The number of all r-combinations
• of a set of n elements denoted
• and read n choose r.
• In the example above, want to find C52,5 .

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Unordered versus ordered selections

• Two ordered selections are the same if
• ? the elements chosen are the same
• ? the elements chosen are in the same order.
• Ordered selections correspond to r-permutations.
  
• The number of all r-permutations is P(n,r) .
• Two unordered selections are the same if
• ? the elements chosen are the same.
• (regardless of the order in which the elements
  are chosen)
• Unordered selections correspond to
  r-combinations.
• The number of all r-combinations is C(n,r) .

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Formula for computing C(n,r)

• Suppose we want to compute P(n,r) .
• Constructing an r-permutation from a set of n
  elements can be thought as a 2-step process
• Step 1 Choose a subset of r elements
• Step 2 Choose an ordering of the r-element
  subset.
• Step 1 can be done in C(n,r) different ways.
• Step 2 can be done in r! different ways.
• (regardless of how the step 1 was performed)
• Based on the multiplication rule, P(n,r) C(n,
  r) r!
• Thus,

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Examples on Combinations

• The number of different 5-card hands
• from a deck of 52 cards
• 4 members from a group of 11
• are supposed to work as a team on a project.
• Q How many distinct 4-person teams can be
  chosen?
• A

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Examples on Combinations

• Suppose in the project example,
• Mary might work on the project only if
  John is also involved in the project.
• Q What is the number of 4-people teams
• that can be chosen from 11 people in this
  case?
• Solution Divide the set S of all possible teams
  into two subsets. Let X be the set of teams with
  Mary
• Y be the set of teams without Mary.
• ? If Mary is in the team, then John is also
  there.
• The remaining 2 spots
• should be filled by the other 9 members.
• Thus, n(X) C(9, 2) 98 / 2 36 .
• ? Suppose Mary is not in the team.
• Then any of the other 10 members can fill
  the 4 spots.
• Thus, n(Y) C(10, 4) (10987) /
  (1234) 210 .
• Based on the addition rule,
• n(S) n(X) n(Y) 36210 246

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Examples on Combinations

• Suppose that 3 cars in a production run of 40 are
  defective.
• A sample of 4 is to be selected to be checked
  for defects.
• Questions
• 1) How many different samples can be chosen?
• 2) How many samples will contain
• exactly one defective car?
• 3) What is the probability that a randomly
  chosen sample will contain exactly one
  defective car?
• 4) How many samples will contain
• at least one defective car?
• Solution
• 1) C(40, 4) (40393837) / (1234)
  91,390

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Examples on Combinations

• 2) How many samples will contain
• exactly one defective car?
• Think of selecting a sample as a 2-step process
• Step 1 Choose the defective cars
• Step 2 Choose the good cars.
• There are C(3,1) ways to choose 1 defective car.
• There are C(37,3) ways to choose 3 good cars.
• By the multiplication rule, the number of
  samples containing exactly 1 defective car
  is
• C(3,1) C(37,3) 3(373635) / (123)
  23,310
• 3) What is the probability that a randomly
  chosen sample will contain exactly one
  defective car?
• The probability 23,310 / 91,390 .255

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Examples on Combinations

• 4) How many samples will contain at least one
  defective car?
• There are 2 ways to answer this question,
• either by the addition rule
• or by the difference rule.
• The solution by the difference rule is less
  intuitive but shorter. Lets solve it by the
  difference rule.
• ( of samples with 1 defective cars)
• (all possible samples)
• ( of samples with no defective cars) .
• But ( of samples with no defective cars)
• ( of samples with only good cars)
  C(37,4)66,045
• Thus, ( of samples with 1 defective cars)
• 91,390 66,045 25,345

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Some Advice about Counting

• Apply the multiplication rule if
• ? The elements to be counted
• can be obtained through a multistep process
• ? Each step is performed
• in a fixed number of ways regardless of
• how preceding steps were performed.
• Apply the addition rule if
• ? The set of elements to be counted
• can be broken up into disjoint subsets.
• Note Often a counting problem is solved by
  applying both the multiplication and addition
  rules (and their variations) at different stages
  of the solution.

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Some Advice about Counting

• In any counting problem, make sure that
• 1) every element is counted
• 2) no element is counted more than once.
• (avoid double counting)
• When using the multiplication rule,
• these directives become
• 1) every outcome should appear as some branch of
  tree
• 2) no outcome should appear
• on more than one branch of tree.
• When using the addition rule, the directives
  become
• 1) every outcome should be in some subset
• 2) the subsets should be disjoint.