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Notes on the Principle of Inclusion-Exclusion (“P.I.E.”) 1 P.I.E. Identities Suppose A1 and A2 and A3 are finite sets, then 1. |A1 ∪ A2 | = |A1 | + |A2 | − |A1 ∩ A2 | 2. |A1 ∪ A2 ∪ A3 | = |A1 | + |A2 | + |A3 | − |A1 ∩ A2 | − |A2 ∩ A3 | − |A2 ∩ A3 | + |A1 ∩ A2 ∩ A3 | More generally, we have the following theorem: P.I.E. Theorem If A1 , A2 , . . . An are any finite sets,then |A1 ∪ A2 ∪ · · · ∪ An | = n X |Ai | − i=1 X |Ai1 ∩ Ai2 | + i1 <i2 X |Ai1 ∩ Ai2 ∩ Ai3 | . . . i1 <i2 <i3 Here is the same theorem with a different notation that might seem less vague (no dots!): P.I.E. Theorem: If A1 , A2 , . . . , An are any finite sets,then n [ X | As | = (−1)|I|+1 |AI |, i=1 I∈Fn where Fn = {I : ∅ ( I ⊆ [n]}, and for any I ⊆ [n], AI = T Ai . i∈I To make sure you understand the notation, lets check that the first P.I.E. identity agrees with the case n = 2. We are summing over the non-empty subsets of [2], namely {1}, {2}, and {1, 2}: | n [ As | = i=1 X (−1)|I|+1 |AI | ∅(I⊆[2] = (1)2 |A{1} | + (−1)2 |A{2} | + (−1)3 |A{1,2} | \ = |A1 | + |A2 | − | Ai | i∈{1,2} = |A1 | + |A2 | − |A1 ∩ A2 |. With the convention that A∅ = U , we have the following important corollary to the P.I.E. Thoerem: Corollary If A1 , A2 , . . . , An are subsets of a finite set U , then \ n [ X n Ai = |U | − | Ai | = (−1)|I| |AI |. i=1 i=1 I⊆[n] Comment: There are 2n terms in this sum, which is a pain in the aneck. However it is often comparatively easy to evaluate the cardinalities |AI |, and sometimes many of the terms are essentially the same because of symmetry. Furthermore, we’ we’ll see later that the alternating sum can be truncated to yield upper and lower bounds that are easier to compute and often convenient. 2 Basic Arithmetic Facts used for the the Examples in the Next Section For any real number x, define bxc = the greatest integer that is less than or equal to x= the integer that is obtained from x by rounding down. For example, b5.257c = 5, b 11 2 c = 5, b3c = 3, bπc = 3, b−πc = −4 . In primary school you learn long division: for any positive integers n and d, there are unique integers q and r such that nd = q + dr , where 0 ≤ r < q. The number q = b nd c is called the quotient, and the number r = n − qd is called the remainder. Note that q is just the number of multiples of d that are less than or equal to n. For example the number of multiples of 7 that are less than or equal to 100 is 14. We can see this three ways: • By listing them (since 14 is small): 7,14,21,28,35,42, 49, 56,63,70,77,84,91,98 • Using long division: divide 7 into 100 and get quotient 14 with remainder 2. • Use a calculator (since n is small or moderate) to find and round down to get 14. 100 7 ≈ 14.2857142857, Another basic fact about numbers: n is divisible by both a and b if and only of n is divisible by the least common multiple of a and b. For example, if a = 360 = 23 · 32 · 5 and b = 525 = 3 · 52 · 7, then the least common multiple of a and b is LCM (360, 525) = 2max(3,0) 3max(2,1) 5max(1,2) 7max(0,1) = 23 · 32 · 52 · 7 = 12600. 3 Examples For each of the following, we are given a set S, and asked to calculate |S| using inclusion exclusion 1. Freshman at Fictitious College are required to take Physics 101 or Chemistry 101. According to the registrar, there are 1000 Freshman, with 550 taking Physics and 600 taking Chemistry. How many students are taking both courses? Answer: By elementary algebra, the first P.I.E. identity is equivalent to |A1 ∩ A2 | = |A| + |A2 | − |A1 ∪ A2 |. (1) Let A1 be the set of freshmen taking Physics 101 and let A2 be the set of freshmen taking Chemistry 101. We are given that |A1 ∪ A2 | = 1000, |A1 | = 550 and |A2 | = 600. Plugging these numbers in into (1), we get |A1 ∩ A2 | = 550 + 600 − 1000 = 150. 2. S = {n ∈ [1000] : n is divisible by 8 or 14} Answer:Let A1 = {n ∈ [1000] : n is divisible by 8 } and A2 = {n ∈ [1000] : n is divisible by 14 }. Then S = A1 ∪ A2 and |S| = |A1 | + |A2 | − |A1 ∩ A2 |. Using the previous section, we get |A1 | = b 1000 8 c = 125 and c = 71. Note that LCM (8, 14) = 56. Therefore |A1 ∩ A2 | = |A2 | = b 1000 14 b 1000 c = 17 and 56 |S| = 125 + 71 − 17 = 179. 3. S = {n ∈ [10000] : n is NOT divisible by any of the numbers 2,5, and 31} Answer Let A1 = {n ∈ [10000] : n is divisible by 2 }, A2 = {n ∈ [10000] : n is divisible by 5 }, A3 = {n ∈ [10000] : n is divisible by 31 }. Then |S| = 10000−|{n ∈ [10000] : n is divisible by at least one of the numbers 2,5, and 31}| = 10000 − |A1 ∪ A2 ∪ A3 | = 10000−(|A1 | + |A2 | + |A3 | − |A1 ∩ A2 | − |A1 ∩ A3 | − |A2 ∩ A3 | + |A1 ∩ A2 ∩ A3 |) = 10000 − b −b 10000 10000 10000 10000 10000 10000 c−b c−b c+b c+b c+b c 2 5 31 10 62 155 10000 c = 10000 − 5000 − 2000 − 322 + 1000 + 161 + 64 − 32 = 3871 310 4. S = the ways to shuffle a deck of cards so that no card stays in its original position. Answer Let U be the set of 52! bijective functions that have [52] as both the domain and the codomain. (These correspond in the obvious way to ways of shuffling: f (i) = j if and only if the card in position i is moved to position j.) Let Ai be the ways to shuffle so that card i stays in its original position, i. More precisely, |Ai | = |{f ∈ U : f (i) = i}| The corollary is now directly applicable. The size of |AI | depends only on |I|. Hence the 52 sets |Ai | all have cardinality 51!, The 52 sets Ai ∩Aj all 2 have cardinality 50!, etc. Since [52] has 52 k subsets of size k,the answer is 52 X 52 |S| = (−1)k+1 (52 − k)! k k=0 Comment For a truly random permutation of the cards, the probability |S| that every card moves is exactly 52! , and this is approximately 1e . Too see this, note that 52 52 k=0 k=0 X (−1)k |S| 1 X 52! = (−1)k+1 (52 − k)! = 52! 52! k!(52 − k)! k! You might remember from calculus that the Taylor-Maclaurin series for x gives ∞ X (−1)k e−1 = . k! k=0