Applied Combinatorics, 4th Ed. Alan Tucker

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Applied Combinatorics, 4th Ed.Alan Tucker

• Section 5.2
• Simple Arrangements and Selections
• Prepared by Patrick
  Asaba and Patrick Scanlon

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Definitions

• Permutation of n distinct objects is an
  arrangement, or ordering, of the n objects.
• r-Permutation of n distinct objects is an
  arrangement using r of the n objects. We use P
  (n, r) to denote the number of r-permutations of
  a set of n objects.
• r-Combination of n distinct objects is an
  unordered selection, or subset, of r out of the n
  objects. We use C (n, r) to denote the number of
  r-combinations of a set of n objects.

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Definitions

• From the multiplication principle we obtain.
• P (n,2) n(n-1),
• P (n,3) n(n-1)(n-2),
• P (n, n) n(n-1)(n-2) x x 3 x 2 x 1
• In enumerating all permutations of n objects, we
  have n choices for the first position in the
  arrangement, n 1 choices (the n 1 remaining
  objects) for the second position,, and finally
  one choice for the last position.

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Definitions

• Using the notation n! n (n 1)(n 2) x 3 x
  2 x 1, we have the formulas
• P (n, n) n! and
• and
• P (n, r) n (n 1)(n 2) x x n (r 1)
  
• This formula can be used to derive a formula for
  C (n, r).
• All r-permutation can be generate by first
  picking r-combination of the n objects and then
  arranging these r objects in any order.
• Thus , and
  solving for C (n, r)
• C (n, r)
• The numbers C (n, r) are frequently called
  binomial coefficients because of their role in
  the binomial expansion

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Examples

• Ranking Wizards
• Ranking is simply an arrangement, or permutation,
  of the n things.

• Arrangements with Repeated Letters

• r would be the letters that are not repeated
• n would be the total number of letters.

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Example 1 Arrangements with Repeated Letters
In the word SYSTEMS how many ways can you arrange
the letters so that all three Ss appear
consecutively?
First, we must find all possible ways for the
letters that are not S (E, M, T, Y). This can be
done by permutation P(7,4) 7!/3! 840
different ways.
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Example 1 contd

• Next, we find out where the Ss go. The trick
  to doing this is to group the Ss as one letter.
  So the possible letters are E, M, T, Y, and SSS.
  These letters can be grouped in 5!, or 120, ways.
• Another approach is to use only one S and add the
  other two in after computing.

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Example 2 Poker Probabilities

• How many 5-card hands can be formed from a
  standard 52-card deck?
• A 5-card hand is a subset of the whole 52-card
  deck, so there are C(52,5) 52! / (47!5!)
  2,598,960 possible 5-card hands.

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Example 2 contd

• What is the probability that a random 5-card hand
  from a standard deck will result in a flush (all
  cards of the same suit)?
• First, find all subsets of 5-cards within the
  same suit C(13,5) 13! / (5!8!) 1287 possible
  flushes for one suit.
• Multiply that result by 4 (for the 4 suits in a
  deck) 1287 x 4 5148 possible flushes.

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Example 2 contd

• To find the probability of getting a flush
  randomly, take the number of possible flushes and
  divide by the total number of possible hands
  5148 / 2,598,960 0.00198, or roughly 0.2

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Example 2 contd

• What is the probability of getting 3, but not 4,
  aces in a 5-card hand?
• First, you must pick 3 of the four aces, done in
  C(4,3) 4 ways.
• Next, fill your hand with any 2 of the 48 other
  cards in the deck C(48,2) 1128 ways.
• Multiply the results to get 1128 x 4 4512 ways
  you can get 3 aces.

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Example 2 contd

• To find the probability, divide by the total
  number of possible hands 4512
  / 2,598,960 0.00174, or roughly 0.17

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Example 3 Forming Committees

• A committee of k people is to be chosen from a
  set of 7 women and 4 men. How many ways can you
  form a committee if
• The committee consists of 3 women and 2 men?
• The committee consists of 4 people, and Mr.
  Baggins must be on it?

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Example 3 contd

• In case 1, you must simply pick between two
  subsets, men and women. Since there are three
  women and 2 men, the numbers are as follows
• C(7,3) x C(4,2) 35 x 6 210 different ways

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Example 3 contd

• In case 2, Mr. Baggins forced entry into the
  committee lessens our choices for who else is on
  the committee. But since there were no other
  restrictions (no gender restrictions), we can
  simply choose 3 from the remaining 10 C(10,3)
  120
• An easy mistake to make is to assume gender
  constrictions read the problems carefully

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Poker Probabilities Example 1

• How many 5-card hands can be formed from a
  standard 52-card deck?
• First we must ask will the hands be ordered or
  unordered. In this example lets suppose we want
  unordered 5-card hands. So it is simply

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Example 2

• If a 5-card hand is chosen at random, what is the
  probability of obtaining a flush (all cards
  having the same suit)?