Simple Arrangements

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Simple Arrangements Selections
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Combinations Permutations

• A permutation of n distinct objects is an
  arrangement, or ordering, of the n objects.
• An r-permutation of n distinct objects is an
  arrangement using r of the n objects.
• A r-combination of n objects is an unordered
  selection, or subset, of r of the n objects.

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Notation

• P(n, r) denotes the number of r-permutations of n
  distinct objects.
• C(n, r) denotes the number of r-combinations of n
  distinct objects.
• It is spoken n choose r.
• The C(n, r) are known as binomial coefficients
  (for reasons that will become clear later).

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Formula for P(n, r)

• By the product principle,
• P(n, 2) n(n - 1) P(n, 3) n(n - 1)(n - 2)
• P(n, n) n!
• n choices for the 1st position
• (n - 1) choices for the 2nd position, ,
• 1 choice for the nth position.
• P(n, r) n(n - 1) . . . (n - (r - 1))
• n(n - 1) . . . (n - (r - 1)) (n -
  r)!/(n - r)!
• n!/(n - r)!

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Formula for C(n, r)

• All r-permutations can be counted by the product
  rule
• 1. pick the r elements to be ordered C(n, r)
• 2. Order the r elements P(r, r) r!
• That is, P(n, r) C(n, r)P(r, r) n!/(n - r)!
• Thus, C(n, r) n!/(r!(n - r)!)
• C(n, r) C(n, n - r).
• Give a counting argument for this.

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Example

• How many ways can 7 women 3 men be arranged in
  a row, if the 3 men must be adjacent?
• Treat the men as a block.
• There are C(8, 1) ways to place the block of men
  among the women.
• There are P(3, 3) ways to arrange the men.
• There are P(7, 7) ways to arrange the women.
• By the product rule, there are (8)(3!)(7!) ways.

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Example

• How many ways are there to arrange the alphabet
  so that there are exactly 5 letters between a
  b?
• Pick the position of the left letter of a b.
  (This forces the position of the other letter.)
• Pick the order of a b 2.
• Arrange the other letters P(24, 24).
• By the product rule C(20, 1)(2)P(24, 24).

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Example

• How many 6-digit numbers without repetition are
  there so that the digits are nonzero, and 1 2
  do not appear consecutively?
• It is simpler to do this indirectly
• There are P(9, 6) ways to arrange 6 nonzero
  digits without repetition.
• Subtract the number of ways to arrange the digits
  so that 1 2 are consecutive

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• There are C(5, 1) ways to pick the leftmost
  position where the 1 or 2 go.
• There are 2 ways to arrange 1 2.
• There are P(7,4) ways to arrange the other
  letters.
• The number of ways to do this is
• P(9, 6) - C(5, 1)(2)P(6, 4) ways to do this.

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Example

• How many ways are there to arrange the letters in
  the word MISSISSIPPI?
• Since the letters are not distinct, the answer is
  less than P(11,11) 11!
• Pick the 1 position where the M goes C(11,1)
• Pick the 4 positions where the Is go C(10,4)
• Pick the 4 positions where the Ss go C(6,4)
• Pick the 2 positions where the Ps go C(2,2)
• There are C(11,1) C(10,4) C(6,4) C(2,2) ways.

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Example

• How many committees of 4 people can be chosen
  from a set of 7 women and 4 men such that there
  are at least 2 women?

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There are at least 2 women?

• If we pick 2 women, then pick 2 more people
  without restriction, we get
• C(7, 2)C(9,2).
• This is wrong certain committees are counted
  more than once.
• See the tree diagram of this use of the product
  rule.

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The Set Composition Rule

• When using the product rule, the elements of each
  component must be distinct.
• In the proposed count, we cannot always tell
  which women were picked in the 1st phase, which
  were picked in the 2nd phase.

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To fix this Use the addition rule

• The set of committees can be partitioned into
• those with 2 women C(7, 2)C(4, 2)
• those with 3 women C(7, 3)C(4, 1)
• those with 4 women C(7, 4)C(4, 0)
• Thus, there are
• C(7, 2)C(4, 2) C(7, 3)C(4, 1) C(7, 4)C(4, 0)
  such committees.

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Example

• How many ways are there to arrange As Us such
  that the 3rd U appears as the 12th letter in a 15
  letter sequence?
• In the 1st 11 letters, U appears exactly 2 times.
• Using the product rule
• Pick the positions of the 1st 2 Us C(11, 2)
• Pick the 3 letters that follow the 12th 23.