1 / 18

Inclusion-Exclusion Formula

Inclusion-Exclusion Formula. S k. Let A 1 , A 2 , …, A n be n sets in a universe U of N elements. Let: S 1 = |A 1 | + |A 2 | + …+ |A n | S 2 = |A 1 A 2 | + |A 1 A 3 | + …+ |A n-1 A n |, i.e., the size of all A i A j for i  j.

rivka
Download Presentation

Inclusion-Exclusion Formula

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Inclusion-Exclusion Formula

  2. Sk • Let A1, A2, …, 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 …Ak+1| + …+ |An-k+1 An-k+2 … An-1An|, i.e., the sizes of all k-ary intersections of the sets. • How many terms are there in S1 , S2 , Sn , Sk?

  3. The Inclusion-Exclusion Formula The # of elements in none of the sets is: | A1A2 …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.

  4. 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.

  5. 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.

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

  7. 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.

  8. Corollary |A1 + A2 +. . .+ An| = S1 - S2 + S3 - S1 - ...+ (-1)n-1Sn. Proof: • A1 + A2 +. . .+ An = U - A1A2 …An • Thus, a count is: N -[N - S1 + S2 - S3 + …+(-1)n Sn] = S1 - S2 + S3 - S1 - ...+ (-1)n-1Sn.

  9. Inclusion-Exclusion Pattern To count the elements of a set that has: property 1andproperty 2and … andproperty 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, A1A2 …Anis the set of all elements that haveproperty 1andproperty 2and … andproperty n.

  10. Inclusion-Exclusion Pattern To count the elements of a set that has: property 1orproperty 2or … orproperty n Define: A1 as the set of elements that haveproperty 1 A2 as the set of elements that haveproperty 2 etc. Then, A1 + A2 +. . .+ An is the set of all elements that haveproperty 1orproperty 2or … orproperty n.

  11. 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 |A1A2A3A4A5A6 | = 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

  12. 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)

  13. 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 |A1A2A3A4| = N - S1 + S2 - S3 + S4

  14. 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.

  15. 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 |A1A2 ... An| = N - S1 + S2 - S3 + … + (-1)n Sn • N = n! • |Ai| = (n-1)!, |AiAj| = (n-2)!, …, |A1A2... An|= 0!

  16. 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)!], C(n,k)(n-k)! = n!/k!. • Thus, the sum equals n! -n!/1! +n!/2! -n!/3! +… + (-1)nn!/n! =

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

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

More Related