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# week7 - S Tanny MAT 344 Spring 1999 N(aN1 aN2 aN3 aN4 = N j...

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S. Tanny MAT 344 Spring, 1999 93 N(a N 1 a N 2 a N 3 a N 4 ) = N - - j N(a i ) % j N(a i a j ) j N(a i a j a k ) Here N = 21 3 (Just put 8 balls in box i and N(a i ) = fill arbitrarily for the 4 % 10 & 1 10 remaining 10) (Put 8 balls in both N(a i a j ) = = i and j, fill arbitrarily from here on) 4 % 2 & 1 2 5 2 N(a i a j a k ) = 0 (3 × 8 = 24 > 18) ° N(a N 1 a N 2 a N 3 a N 4 ) = - 4 + = 246 21 3 13 10 4 2 5 2 Exercise 2 Let A,B be finite sets, * A * = n, * B * = k. Find the number of onto functions f: A 6 B Solution : If n < k, the answer is 0. Assume n \$ k. The number of functions without restriction is k n . For 1 # i # k, let a i denote the property that a function does not have the i th elt of B in its range. Then N(a i ) = (k - 1) n. N(a N i ) counts the functions which do have the i th element of B in their range, and N(a N 1 a N 2 ± a N k ) counts those functions which do have the 1st, 2nd, ± , k th elt of B in their range, i.e. the onto functions. Notice that N(a i a j ) = (k - 2) n while = (k - r) n N a i 1 a i 2 , ± ,a i r Also, the number of r-tuples is . k r Thus, N(a N 1 a N 2 ± a N k ) = k n - (k - 1) n + (k - 2) n - ± k 1 k 2 = j k r 0 ( & 1) r k r k & r n = k!(S( n,k)) / k!(Stirling # of 2nd Kind).

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S. Tanny MAT 344 Spring, 1999 94 Exercise: The number of derangements D n Let a i be the property that i is in its natural position in the permutation B , i.e. B (i) = i. Then D n is the no. of perms with none of the properties a i . Then N = n!, N(a i ) = (n - 1)!, N(a i a j ) = n - 2)! = (n- r)! The number of r-tuples is . Thus, N a i 1 a i 2 , ± ,a i r k r D n = N(a N 1 a N 2 ± a N n ) = j n r 0 ( & 1) r n r (n & r)! Exercise: (Euler Phi Function ) Let n = p e 1 1 p e 2 2 ² p e k k Denote by N (n) the number of integers from 1 to n (incl.) rel. prime to n. Find N (n). Let a i be the property that an integer in [n] is divisible by p i . Then N (n) =N(a N 1 a N 2 ± a N k ). Here N = n, N(a i ) = , etc. n p i , N(a i a j ) n p i p j Thus, N (n) = n - j i n p i % j i 1 ³ j 2 n p i 1 p i 2 & j n p i 1 p i 2 p i 3 ± = n 1 & 1 p 1 1 & 1 p 2 ± 1 & 1 p k Generalization Suppose we want the number of objects which have exactl y m of the properties (any m of the r properties will do, m 0 [0,r]. Let e m be this number. Let s m be the number of objects s m = j N a i 1 a i 2 , ± ,a i m where the sum is taken over all choices of m distinct properties . a i 1 , ± ,a i m N.B. s m counts elements more than once. Every element counted by s m has at least m properties but those with more than m properties get counted many times. For example, if an object has the properties a 1, a 2 , ± , a m + 1 , it gets counted in N(a 1 a 2 ± a m ), N(â 1 a 2 a 3 ± a m + 1 ), N(a 1 â 2 a 3 ± a m + 1 ) etc. 8 8
S. Tanny MAT 344 Spring, 1999 95 “a 1 missing” “a 2 missing” Theorem : e m = s m - m % 1 1 s m % 1 % m % 2 2 s m % 2 & ² ² + (-1) p + (-1) r - m m % p p s m % p % ± % ± m % r & m r & m s r If s 0 = N, this yields the inclusion - exclusion formula for m = 0. Proof : Let’s consider any object x. 1) If it has fewer than m of the properties it contributes 0 to LHS and 0 to each of s m , s m+1 , ± . So 0 to RHS.

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week7 - S Tanny MAT 344 Spring 1999 N(aN1 aN2 aN3 aN4 = N j...

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