InclusionExclusion Formula

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Inclusion-Exclusion Formula
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Sk

• Let A1, A1, , An be n sets in a universe U of N
  elements. Let
• S1 A1 A2 An
• S2 A1 A2 A1 A3 An-1 An, i.e., the
  size of all AiAj for i ? j.
• Sk A1 A2 Ak A1 A2 Ak1 An-k1
  An-k2 An-1An, i.e., the sizes of all k-ary
  intersections of the sets.
• How many terms are there in S1 , S2 , Sn , Sk?

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The Inclusion-Exclusion Formula

• A formula for the of elements in none of the
  sets is
• A1 A2 An N - S1 S2 - S3 (-1)n Sn
• Proof
• We show that the formula counts each element in
• none of the sets once
• 1 or more of the sets a net of 0 times.

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• Case An element that is in none.
• Such an element is added once in the 1st term N
• Since this element is in none of the Ais, it is
  in none of the Si, thus is counted a net of 1
  time.
• Case An element is in exactly 1 of the Ais.
• It is added once in the 1st term, N.
• It is subtracted once in S1.
• It is in no other term, thus is counted a net of
  0.

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• Case An element is in exactly 2 of the Ais.
• It is added once in the 1st term, N.
• It is subtracted twice in S1.
• It is added once in S2.
• It is in no other term, thus is counted a net of
  0 times.

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• Case An element is in exactly k of the Ais.
• It is added once in the 1st term, N.
• It is subtracted C(k, 1) times in S1.
• It is added C(k,2) times in S2.
• It is subtracted C(k,3) times in S3. . . .
• An element in exactly k sets cannot be in any
  intersection of more than k of the Ais.

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• The net count for such an element is
• This binomial sum equals (1 x)k, for x -1.
• Thus, its value is exactly 0.
• In general, the formula counts elements in 1 or
  more of the sets exactly 0 times.

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Corollary

• A1 A2 . . . An S1 - S2 S3 - S1 - ...
  (-1)n-1Sn.
• Proof
• A1 A2 . . . An U - A1 A2 An
• Thus, a count is
• N - N - S1 S2 - S3 (-1)n Sn
• S1 - S2 S3 - S1 - ... (-1)n-1Sn.

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Advice

• To use inclusion-exclusion to count the elements
  of a set that has
• property 1 and property 2 and and property n
• Define
• A1 as the set of elements that do not have
  property 1
• A2 as the set of elements that do not have
  property 2
• etc.
• Then, A1 A2 An is the set of all elements that
  have property 1 and property 2 and and property
  n.

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Advice . . .

• To use inclusion-exclusion to count the elements
  of a set that has
• property 1 or property 2 or or property n
• Define
• A1 as the set of elements that have property 1
• A2 as the set of elements that have property 2,
  etc.
• Then, A1 A2 . . . An is the set of all
  elements that have property 1 or property 2 or
  or property n.

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Example 1

• How many ways are there to roll 10 distinct dice
  so that all 6 faces appear?
• Let Ai be the set of ways that face i does not
  appear.
• Then, we want A1 A2 A3 A4 A5 A6
• N - S1 S2 - S3 S4 - S5 S6
• 610 - C(6,1)510 C(6,2)410 C(6,3)310
  C(6,4)210 C(6,5)110

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Example 2

• What is the probability that a 10-card hand has
  at least one 4-of-a-kind?
• Let A1 be the set of all 10-card hands with 4
  aces.
• Let AK be the set of all 10-card hands with 4
  kings.
• Then, we want A1 A2 . . . AK
• S1 - S2 S3 - . . . S13 C(13,1)C(48,6) -
  C(13,2)C(44,2)
• The probability C(13,1)C(48,6) -
  C(13,2)C(44,2) / C(52,10)

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Example 3

• How many integer solutions of
• x1 x2 x3 x4 30 are there with
• 0 ? xi, x1 ? 5, x2 ? 10, x3 ? 15, x4 ? 21 ?
• Let A1 be the set of solutions where x1 ? 6.
• Let A2 be the set of solutions where x2 ? 11.
• Let A3 be the set of solutions where x3 ? 16.
• Let A4 be the set of solutions where x4 ? 22.
• We want A1 A2 A3 A4 N - S1 S2 - S3 S4

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• N C(30 4 - 1, 4 - 1)
• A1 C(30 - 6 4 - 1, 4 - 1)
• A2 C(30 - 11 4 - 1, 4 - 1)
• A3 C(30 - 16 4 - 1, 4 - 1)
• A4 C(30 - 22 4 - 1, 4 - 1)
• A1 A2 C(30 - 17 4 - 1, 4 - 1)
• A1 A3 C(30 - 22 4 - 1, 4 - 1)
• A1 A4 C(30 - 28 4 - 1, 4 - 1)
• A2 A3 C(30 - 27 4 - 1, 4 - 1)
• A2 A4 0 A3 A4
• All 3-intersections 4-intersections have 0
  elements in them.

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Derangements

• What is the probability that if n people randomly
  reach into a dark closet to retrieve their hats,
  no person receives their own hat?
• Let Ai be the set outcomes where person i
  receives his/her own hat.
• We need A1 A2 ... An N - S1 S2 - S3
  (-1)n Sn
• N n!
• Ai (n-1)!, AiAj (n-2)!, , A1A2... An
  0!

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• N - S1 S2 - S3 (-1)n Sn
• n! - C(n,1)(n-1)! C(n,2)(n-2)! - (-1)n
  C(n,n)0!
• Since C(n,k) n!/k!(n-k)!, in the sum above
  the term C(n,k) (n-k)! n!/k!.
• Thus, the sum equals
• n! - n!/1! n!/2! - n!/3! (-1)n n!/n!

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• This sum is the number of good outcomes.
• The probability, then, is this number of all
  possible (n!) outcomes

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• This is the 1st n terms of the power series for
• This power series converges quickly.
• The numerator, the number of permutations that
  leave no element fixed, is denoted Dn, and is
  called the number of derangements of n elements.

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