Benjamin, Arthur T.

Bennet, Curtis D.

Newberger, Florence

Scholarship@Claremont

Claremont CollegesScholarship @ ClaremontAll HMC Faculty Publications and Research HMC Faculty Scholarship12-1-2005Recounting the Odds of an Even DerangementArthur T. BenjaminHarvey Mudd CollegeCurtis D. BennetLoyola Marymount UniversityFlorence NewbergerCalifornia State University - Long BeachThis Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusionin All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contactscholarship@cuc.claremont.edu.Recommended CitationBenjamin, Arthur T., Curtis D. Bennet and Florence Newberger. "Recounting the Odds of an Even Derangement." MathematicsMagazine, Vol 78, No. 5, pp. 387-390, December 2005.Recounting the Odds of an Even DerangementAuthor(s): Arthur T. Benjamin, Curtis T. Bennett and Florence NewbergerSource: Mathematics Magazine, Vol. 78, No. 5 (Dec., 2005), pp. 387-390Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/30044197 .Accessed: 11/06/2013 18:35Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp .JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org. .Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toMathematics Magazine.http://www.jstor.org This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 18:35:08 PMAll use subject to JSTOR Terms and ConditionsVOL. 78, NO. 5, DECEMBER 2005 387 We see the potential value of this proof as twofold. First, it appears cleaner and shorter than what is found in most texts. And our constant, ln(8), is modest com- pared to Sierpinski's 4 [7], Apostol's 6 [1], or the 32 ln(2) offered in earlier editions of Niven and Zuckerman [6]. LeVeque [5], Hardy and Wright [4], and the latest edition of Niven and Zuckerman [6] give no particular constant, merely proving that one ex- ists. Chebychev [3] achieved a much smaller constant han ours, but with considerably more effort. We hope that our short proof will be found to have pedagogical value. REFERENCES 1. Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976. 2. Bela Bollobas, Erd6s first paper: A proof of Bertrand's Postulate, pre-print, 1998. 3. P.L. Chebychev, Memoire sur les nombres premiers, J. Math. Pures et Appl. 17, (1852), 366-390. 4. G.H. Hardy and E.M. Wright, An Introduction tothe Theory ofNumbers, 5th Edition, Clarendon Press, Oxford, 1979. 5. William J. LeVeque, Fundamentals of Number Theory, Dover Publications, New York, 1977. 6. I. Niven and H.S. Zuckerman, An Introduction tothe Theory of Numbers, 3rd, 4th, 5th Editions, Wiley, New York, 1972, 1980, 1991. 7. Waclaw Sierpinski, Elementary Theory of Numbers, PWN, Warsaw, 1964. Recounting the Odds of an Even Derangement ARTHUR T. BENJAMIN Harvey Mudd College Claremont, CA 91711 benjamin@hmc.edu CURTIS T. BENNETT Loyola Marymount University Los Angeles, CA 90045 cbennett@lmu.edu FLORENCE NEWBERGER California State University Long Beach, CA 90840-1001 fnewberg@csulb.edu Odd as it may sound, when n exams are randomly returned to n students, the probabil- ity that no student receives his or her own exam is almost exactly 1/e (approximately 0.368), for all n > 4. We call a permutation with no fixed points, a derangement, and we let D(n) denote the number of derangements of n elements. For n > 1, it can be shown that D(n) = -_,(-1)kn!/k!, and hence the odds that a random permutation of n elements has no fixed points is D(n)/n!, which is within 1/(n + 1)! of 1/e [1]. Permutations come in two varieties: even and odd. A permutation is even if it can be achieved by making an even number of swaps; otherwise it is odd. Thus, one might even be interested to know that if we let E (n) and O (n) respectively denote the number of even and odd derangements of n elements, then (oddly enough), D(n) + (n - 1)(-1)n-1 E (n) = 2 This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 18:35:08 PMAll use subject to JSTOR Terms and Conditions388 MATHEMATICS MAGAZINE and D(n) - (n - 1)(-1)n-1 O (n) = 2 The above formulas are an immediate consequence of the equation E (n) + O (n) = D(n), which is obvious, and the following theorem, which is the focus of this note. THEOREM. For n > 1, E(n) - O(n) = (-1)n"-(n - 1). (1) Proof 1: Determining a Determinant The fastest way to derive equation (1), as is done in [3], is to compute a determinant. Recall that an n-by-n matrix A = [aij]nj=l has determinant det(A) = 7rESn alr(I)a27z(2) ...a (n) sgn(n), (2) where Sn is the set of all permutations of 1, ..., n}, sgn(r) = 1 when 7r is even, and sgn(7r) = -1 when 7r is odd. Let An denote the n-by-n matrix whose nondiagonal entries are aij = 1 (for i 0 j), with zeroes on the diagonal. For example, when n = 4, a By (2), every permutation that is not a derangement will contribute 0 to the sum (since it uses at least one of the diagonal entries), every even derangement will con- tribute 1 to the sum, and every odd derangement will contribute -1 I to the sum. Conse- quently, det(An) = E(n) - O(n). To see that det(An) = (-1)"-'(n - 1), observe that A, = Jn - In, where Jn is the matrix of all ones and In is the identity matrix. Since Jn has rank one, zero is an eigenvalue of J,, with multiplicity n - 1, and its other eigen- value is n (with an eigenvector of all is). Apply Jn - In to the eigenvectors of J, to find the eigenvalues of An: -1 with multiplicity n - 1 and n - 1 with multiplicity 1. Multiplying the eigenvalues gives us det(An) = (-1)n-1 (n - 1), as desired. 0 A 1996 Note in the MAGAZINE [2] gave even odder ways to determine the deter- minant of An . Although the proof by determinants is quick, the form of (1) suggests that there should also exist an almost one-to-one correspondence between the set of even de- rangements and the set of odd derangements. Proof 2: Involving an Involution Let D, denote the set of derangements of {1,..., n}, and let X, be a set of n - 1 exceptional derangements (that we spec- ify later), each with sign (-1)"-1. We exhibit a sign reversing involution on Dn - X,. That is, letting T, = D, - X,, we find an invertible function f : T, --+ T, such that for ,r in T,, r and f (r) have opposite signs, and f(f (r)) = xr. In other words, ex- cept for the n - 1 exceptional derangements, every even derangement "holds hands" with an odd derangement, and vice versa. From this, it immediately follows that BN Before describing f, we establish some notation. We express each r in D, as the product of k disjoint cycles C1, ..., Ck with respective lengths mi, ..., mk for some This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 18:35:08 PMAll use subject to JSTOR Terms and ConditionsVOL. 78, NO. 5, DECEMBER 2005 389 k > 1. We follow the convention that each cycle begins with its smallest element, and the cycles are listed from left to right in increasing order of the first element. In particular, C1 = (1 a2 ... aml) and, if k > 2, C2 begins with the smallest element that does not appear in C1. Since ir is a derangement on n elements, we must have mi > 2 for all i, and Lk, mi = n. Finally, since a cycle of length m has sign (-1)m-1, it follows that r has sign (-1)i=, (mi 1) - (-1)n~k Let 7r be a derangement in Dn with first cycle C1 = (1 a2 ... am) for some m > 2. We say that 7r has extraction point e > 2 if e is the smallest number in the set {2, ...,n} - {a2} for which C1 does not end with the numbers of {2,..., e} - {a2} written in decreasing order. Note that 7r will have extraction point e = 2 unless the number 2 appears as the second term or last term of C1. We illustrate this definition with some pairs of examples from D9. Notice that in each pair below, the number of cycles of 7 and 7r' differ by one, and the extraction point e occurs in the first cycle of 7r and is the leading element of the second cycle of 7'. r = (1 9 7 2 8)(3 6)(4 5) and 7r' = (19 7)(2 8)(3 6)(4 5) have e = 2. 7r = (1297385)(46) and r' = (1297)(385)(46) have e = 3. 7r= (197385 2)(46) and 7' = (1972)(385)(46) have e = 3. 7r= (1948532)(67) and 7r' = (1932)(485)(67) have e = 4. 7r = (1495832)(67) and 7r' = (14932)(58)(67) have e = 5. 7r= (138697542) and r'= (138542)(697) have e = 6. Observe that every derangement r in Dn contains an extraction point unless r consists of a single cycle of the form 7 = (1 a2 Z), where Z is the ordered set {2, 3, ...., n - 1, n} - {a2}, written in decreasing order. For example, the 9-element derangement (1 5 9 8 7 6 4 3 2) has no extraction point. Since a2 can be any element of {2, ... , n}, there are exactly n - 1 derangements of this type, all of which have sign (-1)n-1. We let X, denote the set of derangements of this form. Our problem reduces to finding a sign reversing involution f over T, = D, - X,. Suppose nr in Tn has extraction point e. Then the first cycle C1 of r ends with the (possibly empty) ordered subset Z consisting of the elements of {2, .. , e - 11 - {a2} written in decreasing order. Our sign reversing involution f : Tn -> Tn can then be succinctly described as follows: f (1 a2 X e Y Z)a (1 a2 X Z)(e Y)a, (3) where X and Y are ordered subsets, Y is nonempty, and a is the rest of the derange- ment r. Notice that since the number of cycles of 7r and f(7r) differ by one, they must be of opposite signs. The derangements on the left side of (3) are those for which the extraction point e is in the first cycle. In this case, Y must be nonempty, since otherwise "e Z" would be a longer decreasing sequence and e would not be the extraction point. The derangements on the right side of (3) are those for which the extraction point e is not in the first cycle (and must therefore be the leading element of the second cycle). In this case, Y is nonempty since iv is a derangement. Thus for any derangement i, the derangement f (i) is also written in standard form, with the same extraction point e and with the same associated ordered subset Z. Another way to see that n and f (i) have opposite signs is to notice that f(iv) = (xy)i (multiplying from left to right), where x is the last element of X (x = a2 when X is empty), and y is the last element This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 18:35:08 PMAll use subject to JSTOR Terms and Conditions390 MATHEMATICS MAGAZINE of Y. Either way, f (f (r)) = 7r, and f is a well-defined, sign-reversing involution, as desired. I In summary, we have shown combinatorially that for all values of n, there are almost as many even derangements as odd derangements of n elements. Or to put it another way, when randomly choosing a derangement with at least five elements, the odds of having an even derangement are nearly even. Acknowledgment. We are indebted to Don Rawlings for bringing this problem to our attention and we thank Magnhild Lien, Will Murray, and the referees for many helpful ideas. REFERENCES 1. R. A. Brualdi, Introductory Combinatorics, 3rd ed., Prentice-Hall, New Jersey, 1999. 2. C. A. McCarthy and A. T. Benjamin, Determinants of the tournaments, this MAGAZINE, 69 (1996), 133-135. 3. C. D. Olds, Odd and even derangements, Solution E907, Amer Math. Monthly, 57 (1950), 687-688. Volumes of Generalized Unit Balls XIANFU WANG UBC Okanagan 3333 University Way, Kelowna B.C., Canada, V1V 1V7 shawn.wang@ubc.ca Diamonds, cylinders, squares, stars, and balls. These geometric figures are familiar to undergraduate students, but what could they possibly have in common? One answer is: They are generalized balls. The standard Euclidean ball can be distorted into a variety of strange-shaped balls by linear and nonlinear transformations. The purpose of this note is to give a unified formula for computing the volumes of generalized unit balls in n-dimensional spaces. A generalized unit ball in Rn is described by the set BPI (1) where pl > 0, P2 > 0,..., Pn > 0. When the numbers pi1 ...., pn are all greater than or equal to 1, the unit ball B is convex. Since IxlP is not concave on [-1, 1] for 0 < p < 1, BP,...Pn is not neces- sarily convex anymore when n > 1. When p1 = P2 = ...= Pn, = p > 1, we obtain the usual 1, ball. The 12 ball is denoted by B. By choosing different numbers pi, we can alter the appearance of the generalized balls greatly, as shown in FIGURE 1 with examples in R3 Motivated by an article by Folland [5], I derived a unified formula for calculating the volume of these balls. Although the volume formulas for the standard Euclidean ball TB and simplex have been known for a long time [4, pp. 208, 220], the unified formula is (relatively) new. It is surprising that no matter how strange the balls look, the volume of any ball can be computed by a single formula, as follows: THEOREM. Assume pi, ..., pn > O. The volume of the unit ball Bp is equal to This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 18:35:08 PMAll use subject to JSTOR Terms and Conditions

Recounting the Odds of an Even Derangement

Bennett, Curtis T.

Loyola Marymount University

Digital Commons@Loyola Marymount Universityand Loyola Law SchoolMathematics Faculty Works Mathematics12-1-2005Recounting the Odds of an Even DerangementArthur T. BenjaminHarvey Mudd CollegeCurtis T. BennettLoyola Marymount University, cbennett@lmu.eduFlorence NewbergerCalifornia State University, Long BeachThis Article is brought to you for free and open access by the Mathematics at Digital Commons @ Loyola Marymount University and Loyola LawSchool. It has been accepted for inclusion in Mathematics Faculty Works by an authorized administrator of Digital Commons@Loyola MarymountUniversity and Loyola Law School. For more information, please contact digitalcommons@lmu.edu.Repository CitationBenjamin, Arthur T.; Bennett, Curtis T.; and Newberger, Florence, "Recounting the Odds of an Even Derangement" (2005).Mathematics Faculty Works. 108.http://digitalcommons.lmu.edu/math_fac/108Recommended CitationBenjamin, Arthur T., Curtis T. Bennett, and Florence Newberger. "Recounting the Odds of an Even Derangement." MathematicsMagazine 78, no. 5 (2005): 387-90. doi:10.2307/30044197. VOL. 78, NO. 5, DECEMBER 2005 387 We see the potential value of this proof as twofold. First, it appears cleaner and shorter than what is found in most texts. And our constant, ln(8), is modest com- pared to Sierpinski's 4 [7], Apostol's 6 [1], or the 32 ln(2) offered in earlier editions of Niven and Zuckerman [6]. LeVeque [5], Hardy and Wright [4], and the latest edition of Niven and Zuckerman [6] give no particular constant, merely proving that one ex- ists. Chebychev [3] achieved a much smaller constant than ours, but with considerably more effort. We hope that our short proof will be found to have pedagogical value. REFERENCES 1. Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976. 2. Bela Bollobas, Erd6s first paper: A proof of Bertrand's Postulate, pre-print, 1998. 3. P.L. Chebychev, Memoire sur les nombres premiers, J. Math. Pures et Appl. 17, (1852), 366-390. 4. G.H. Hardy and E.M. Wright, An Introduction to the Theory ofNumbers, 5th Edition, Clarendon Press, Oxford, 1979. 5. William J. LeVeque, Fundamentals of Number Theory, Dover Publications, New York, 1977. 6. I. Niven and H.S. Zuckerman, An Introduction to the Theory of Numbers, 3rd, 4th, 5th Editions, Wiley, New York, 1972, 1980, 1991. 7. Waclaw Sierpinski, Elementary Theory of Numbers, PWN, Warsaw, 1964. Recounting the Odds of an Even Derangement ARTHUR T. BENJAMIN Harvey Mudd College Claremont, CA 91711 benjamin@hmc.edu CURTIS T. BENNETT Loyola Marymount University Los Angeles, CA 90045 cbennett@lmu.edu FLORENCE NEWBERGER California State University Long Beach, CA 90840-1001 fnewberg@csulb.edu Odd as it may sound, when n exams are randomly returned to n students, the probabil- ity that no student receives his or her own exam is almost exactly 1/e (approximately 0.368), for all n > 4. We call a permutation with no fixed points, a derangement, and we let D(n) denote the number of derangements of n elements. For n > 1, it can be shown that D(n) = -_,(-1)kn!/k!, and hence the odds that a random permutation of n elements has no fixed points is D(n)/n!, which is within 1/(n + 1)! of 1/e [1]. Permutations come in two varieties: even and odd. A permutation is even if it can be achieved by making an even number of swaps; otherwise it is odd. Thus, one might even be interested to know that if we let E (n) and O (n) respectively denote the number of even and odd derangements of n elements, then (oddly enough), D(n) + (n - 1)(-1)n-1 E (n) =  2 388 MATHEMATICS MAGAZINE and D(n) - (n - 1)(-1)n-1 O (n) = 2 The above formulas are an immediate consequence of the equation E (n) + O (n) = D(n), which is obvious, and the following theorem, which is the focus of this note. THEOREM. For n > 1, E(n) - O(n) = (-1)n"-(n - 1). (1) Proof 1: Determining a Determinant The fastest way to derive equation (1), as is done in [3], is to compute a determinant. Recall that an n-by-n matrix A = [aij]nj=l has determinant det(A) = 7rESn alr(I)a27z(2) ...a (n) sgn(n), (2) where Sn is the set of all permutations of 1, ..., n}, sgn(r) = 1 when 7r is even, and sgn(7r) = -1 when 7r is odd. Let An denote the n-by-n matrix whose nondiagonal entries are aij = 1 (for i 0 j), with zeroes on the diagonal. For example, when n = 4, a By (2), every permutation that is not a derangement will contribute 0 to the sum (since it uses at least one of the diagonal e tries), every even derangement will con- tribute 1 to the sum, and every odd derangement will contribute -1 I to the sum. Conse- quently, det(An) = E(n) - O(n). To see that det(An) = (-1)"-'(n - 1), observe that A, = Jn - In, where Jn is the matrix of all ones and I  is the identity matrix. Since Jn has rank one, zero is a  eigenvalue of J,, with multiplicity n - 1, and its other eigen- value is n (with an eigenvector of all is). Apply Jn - In to the eigenvectors of J, to find the eigenvalues of An: -1 with multiplicity n - 1 and n - 1 with multiplicity 1. Multiplying the eigenvalues gives us det(An) = (-1)n-  (n - 1), as desired. 0 A 1996 Note in the MAGAZINE [2] gave even odder ways to determine the deter- minant of An . Although the proof by determinants is quick, the form of (1) suggests that there should also exist an almost one-to-one correspondence between the set of even de- rangements and the set of odd derangements. Proof 2: Involving an Involution Let D, denote the set of derangements of {1,..., n}, a d let X, be a set of n - 1 exceptional derangements (that we spec- ify later), each with sign (-1)"-1. We exhibit a sign reversing involution on Dn - X,. That is, letting T, = D, - X,, we find an invertible function f : T, --+ T, such that for ,r in T,, r and f (r) have opposite sig s, and f(  (r)) = xr. In other words, ex- cept for the n - 1 exceptional derangements, every even derangement "holds hands" with an odd derangement, and vice versa. From this, it immediately follows that BN efore describing f, we establish some notation. We express each r in D, as the product of k disjoint cycles C1, ..., Ck with respective lengths mi, ..., mk for some VOL. 78, NO. 5, DECEMBER 2005 389 k > 1. We follow the convention that each cycle begins with its smallest element, and the cycles are listed from left to right in increasing order of the first element. In particular, C1 = (1 a2 ... aml) and, if k > 2, C2 begins with the smallest element that does not appear in C1. Since ir is a derangement on n elements, we must have mi > 2 for all i, and Lk, mi = n. Finally, since a cycle of length m has sign (-1)m-1, it follows that r has sign (-1)i=, (mi 1) - (-1)n~k Let 7r be a derangement in Dn with first cycle C1 = (1 a2 ... am) for some m > 2. We say that 7r has extraction point e > 2 if e is the smallest number in the set {2, ...,n} - {a2} for which C1 does not end with the numbers of {2,..., e} - {a2} written in decreasing order. Note that 7r will have extraction point e = 2 unless the number 2 appears as the second term or last term of C1. We illustrate this definition with some pairs of examples from D9. Notice that in each pair below, the number of cycles of 7 and 7r' differ by one, and the extraction point e occurs in the first cycle of 7r and is the leading element of the second cycle of 7'. r = (1 9 7 2 8)(3 6)(4 5) and 7r' = (19 7)(2 8)(3 6)(4 5) have e = 2. 7r = (1297385)(46) and r' = (1297)(385)(46) have e = 3. 7r= (197385 2)(46) and 7' = (1972)(385)(46) have e = 3. 7r= (1948532)(67) and 7r' = (1932)(485)(67) have e = 4. 7r = (1495832)(67) and 7r' = (14932)(58)(67) have e = 5. 7r= (138697542) and r'= (138542)(697) have e = 6. Observe that every derangement r in Dn contains an extraction point unless r consists of a single cycle of the form 7 = (1 a2 Z), where Z is the ordered set {2, 3, ...., n - 1, n} - {a2}, written in decreasing order. For example, the 9-element derangement (1 5 9 8 7 6 4 3 2) has no extraction point. Since a2 can be any element of {2, ... , n}, there are exactly n - 1 derangements of this type, all of which have sign (-1)n-1. We let X, denote the set of derangements of this form. Our problem reduces to finding a sign reversing involution f over T, = D, - X,. Suppose nr in Tn has extraction point e. Then the first cycle C1 of r ends with the (possibly empty) ordered subset Z consisting of the elements of {2, .. , e - 11 - {a2} written in decreasing order. Our sign reversing involution f : Tn -> Tn can then be succinctly described as follows: f (1 a2 X e Y Z)a (1 a2 X Z)(e Y)a, (3) where X and Y are ordered subsets, Y is nonempty, and a is the rest of the derange- ment r. Notice that since the number of cycles of 7r and f(7r) differ by one, they must be of opposite signs. The derangements on the left side of (3) are those for which the extraction point e is in the first cycle. In this case, Y must be nonempty, since otherwise "e Z" would be a longer decreasing sequence and e would not be the extraction point. The derangements on the right side of (3) are those for which the extraction point e is not in the first cycle (and must therefore be the leading element of the second cycle). In this case, Y is nonempty since iv is a derangement. Thus for any derangement i, the derangement f (i) is also written in standard form, with the same extraction point e and with the same associated ordered subset Z. Another way to see that n and f (i) have opposite signs is to notice that f(iv) = (xy)i (multiplying from left to right), where x is the last element of X (x = a2 when X is empty), and y is the last element 390 MATHEMATICS MAGAZINE of Y. Either way, f (f (r)) = 7r, and f is a well-defined, sign-reversing involution, as desired. I In summary, we have shown combinatorially that for all values of n, there are almost as many even derangements as odd derangements of n elements. Or to put it another way, when randomly choosing a derangement with at least five elements, the odds of having an even derangement are nearly even. Acknowledgment. We are indebted to Don Rawlings for bringing this problem to our attention and we thank Magnhild Lien, Will Murray, and the referees for many helpful ideas. REFERENCES 1. R. A. Brualdi, Introductory Combinatorics, 3rd ed., Prentice-Hall, New Jersey, 1999. 2. C. A. McCarthy and A. T. Benjamin, Determinants of the tournaments, this MAGAZINE, 69 (1996), 133-135. 3. C. D. Olds, Odd and even derangements, Solution E907, Amer Math. Monthly, 57 (1950), 687-688. Volumes of Generalized Unit Balls XIANFU WANG UBC Okanagan 3333 University Way, Kelowna B.C., Canada, V1V 1V7 shawn.wang@ubc.ca Diamonds, cylinders, squares, stars, and balls. These geometric figures are familiar to undergraduate students, but what could they possibly have in common? One answer is: They are generalized balls. The standard Euclidean ball can be distorted into a variety of strange-shaped balls by linear and nonlinear transformations. The purpose of this note is to give a unified formula for computing the volumes of generalized unit balls in n-dimensional spaces. A generalized unit ball in Rn is described by the set BPI (1) where pl > 0, P2 > 0,..., Pn > 0. When the numbers pi1 ...., pn are all greater than or equal to 1, the unit ball B is convex. Since IxlP is not concave on [-1, 1] for 0 < p < 1, BP,...Pn is not neces- sarily convex anymore when  > 1. When p1 = P2 = ...= Pn, = p > 1, we obtain the usual 1, ball. The 12 ball is denoted by B. By choosing different numbers pi, we can alter the appearance of the generalized balls greatly, as shown in FIGURE 1 with examples in R3 Motivated by an article by Folland [5], I derived a unified formula for calculating the volume of these balls. Although the volume formulas for the standard Euclidean ball TB and simplex have been known for a long time [4, pp. 208, 220], the unified formula is (relatively) new. It is surprising that no matter how strange the balls look, the volume of any ball can be computed by a single formula, as follows: THEOREM. Assume pi, ..., pn > O. The volume of the unit ball Bp is equal to

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All HMC Faculty Publications and Research HMC Faculty Scholarship
12-1-2005
Recounting the Odds of an Even Derangement
Arthur T. Benjamin
Harvey Mudd College
Curtis D. Bennet
Loyola Marymount University
Florence Newberger
California State University - Long Beach
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Benjamin, Arthur T., Curtis D. Bennet and Florence Newberger. "Recounting the Odds of an Even Derangement." Mathematics
Magazine, Vol 78, No. 5, pp. 387-390, December 2005.
Recounting the Odds of an Even Derangement
Author(s): Arthur T. Benjamin, Curtis T. Bennett and Florence Newberger
Source: Mathematics Magazine, Vol. 78, No. 5 (Dec., 2005), pp. 387-390
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VOL. 78, NO. 5, DECEMBER 2005 387 
We see the potential value of this proof as twofold. First, it appears cleaner and 
shorter than what is found in most texts. And our constant, ln(8), is modest com- 
pared to Sierpinski's 4 [7], Apostol's 6 [1], or the 32 ln(2) offered in earlier editions of 
Niven and Zuckerman [6]. LeVeque [5], Hardy and Wright [4], and the latest edition 
of Niven and Zuckerman [6] give no particular constant, merely proving that one ex- 
ists. Chebychev [3] achieved a much smaller constant han ours, but with considerably 
more effort. We hope that our short proof will be found to have pedagogical value. 
REFERENCES 
1. Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976. 
2. Bela Bollobas, Erd6s first paper: A proof of Bertrand's Postulate, pre-print, 1998. 
3. P.L. Chebychev, Memoire sur les nombres premiers, J. Math. Pures et Appl. 17, (1852), 366-390. 
4. G.H. Hardy and E.M. Wright, An Introduction tothe Theory ofNumbers, 5th Edition, Clarendon Press, Oxford, 
1979. 
5. William J. LeVeque, Fundamentals of Number Theory, Dover Publications, New York, 1977. 
6. I. Niven and H.S. Zuckerman, An Introduction tothe Theory of Numbers, 3rd, 4th, 5th Editions, Wiley, New 
York, 1972, 1980, 1991. 
7. Waclaw Sierpinski, Elementary Theory of Numbers, PWN, Warsaw, 1964. 
Recounting the Odds 
of an Even Derangement 
ARTHUR T. BENJAMIN 
Harvey Mudd College 
Claremont, CA 91711 
benjamin@hmc.edu 
CURTIS T. BENNETT 
Loyola Marymount University 
Los Angeles, CA 90045 
cbennett@lmu.edu 
FLORENCE NEWBERGER 
California State University 
Long Beach, CA 90840-1001 
fnewberg@csulb.edu 
Odd as it may sound, when n exams are randomly returned to n students, the probabil- 
ity that no student receives his or her own exam is almost exactly 1/e (approximately 
0.368), for all n > 4. We call a permutation with no fixed points, a derangement, and 
we let D(n) denote the number of derangements of n elements. For n > 1, it can be 
shown that D(n) = 
-_,(-1)kn!/k!, and hence the odds that a random permutation 
of n elements has no fixed points is D(n)/n!, which is within 1/(n + 1)! of 1/e [1]. 
Permutations come in two varieties: even and odd. A permutation is even if it can 
be achieved by making an even number of swaps; otherwise it is odd. Thus, one might 
even be interested to know that if we let E (n) and O (n) respectively denote the number 
of even and odd derangements of n elements, then (oddly enough), 
D(n) + (n - 1)(-1)n-1 E (n) = 2 
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388 MATHEMATICS MAGAZINE 
and 
D(n) - (n - 1)(-1)n-1 O (n) = 2 
The above formulas are an immediate consequence of the equation E (n) + O (n) = 
D(n), which is obvious, and the following theorem, which is the focus of this note. 
THEOREM. For n > 1, 
E(n) - O(n) = (-1)n"-(n - 1). (1) 
Proof 1: Determining a Determinant The fastest way to derive equation (1), as is 
done in [3], is to compute a determinant. Recall that an n-by-n matrix A = [aij]nj=l 
has determinant 
det(A) = 
7rESn alr(I)a27z(2) ...a (n) sgn(n), (2) 
where Sn is the set of all permutations of 1, ..., n}, sgn(r) = 1 when 7r is even, and 
sgn(7r) = -1 when 7r is odd. Let An denote the n-by-n matrix whose nondiagonal 
entries are aij = 1 (for i 0 j), with zeroes on the diagonal. For example, when n = 4, 
a 
By (2), every permutation that is not a derangement will contribute 0 to the sum 
(since it uses at least one of the diagonal entries), every even derangement will con- 
tribute 1 to the sum, and every odd derangement will contribute -1 I to the sum. Conse- 
quently, det(An) = E(n) - O(n). To see that det(An) = (-1)"-'(n - 1), observe that 
A, = Jn - In, where Jn is the matrix of all ones and In is the identity matrix. Since Jn 
has rank one, zero is an eigenvalue of J,, with multiplicity n - 1, and its other eigen- 
value is n (with an eigenvector of all is). Apply Jn - In to the eigenvectors of J, to 
find the eigenvalues of An: -1 with multiplicity n - 1 and n - 1 with multiplicity 1. 
Multiplying the eigenvalues gives us det(An) = (-1)n-1 (n - 1), as desired. 0 
A 1996 Note in the MAGAZINE [2] gave even odder ways to determine the deter- 
minant of An . 
Although the proof by determinants is quick, the form of (1) suggests that there 
should also exist an almost one-to-one correspondence between the set of even de- 
rangements and the set of odd derangements. 
Proof 2: Involving an Involution Let D, denote the set of derangements of 
{1,..., n}, and let X, be a set of n - 1 exceptional derangements (that we spec- 
ify later), each with sign (-1)"-1. We exhibit a sign reversing involution on Dn - X,. 
That is, letting T, = D, - X,, we find an invertible function f : T, --+ T, such that 
for ,r in T,, r and f (r) have opposite signs, and f(f (r)) = xr. In other words, ex- 
cept for the n - 1 exceptional derangements, every even derangement "holds hands" 
with an odd derangement, and vice versa. From this, it immediately follows that 
BN 
Before describing f, we establish some notation. We express each r in D, as the 
product of k disjoint cycles C1, ..., Ck with respective lengths mi, ..., mk for some 
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VOL. 78, NO. 5, DECEMBER 2005 389 
k > 1. We follow the convention that each cycle begins with its smallest element, 
and the cycles are listed from left to right in increasing order of the first element. In 
particular, C1 = (1 a2 ... aml) and, if k > 2, C2 begins with the smallest element that does not appear in C1. Since ir is a derangement on n elements, we must have mi > 2 
for all i, and Lk, mi = n. Finally, since a cycle of length m has sign (-1)m-1, it 
follows that r has sign (-1)i=, (mi 1) - (-1)n~k 
Let 7r be a derangement in Dn with first cycle C1 = (1 a2 ... am) for some 
m > 2. We say that 7r has extraction point e > 2 if e is the smallest number in the set 
{2, ...,n} - {a2} for which C1 does not end with the numbers of {2,..., e} - {a2} 
written in decreasing order. Note that 7r will have extraction point e = 2 unless the 
number 2 appears as the second term or last term of C1. We illustrate this definition 
with some pairs of examples from D9. Notice that in each pair below, the number of 
cycles of 7 and 7r' differ by one, and the extraction point e occurs in the first cycle of 
7r and is the leading element of the second cycle of 7'. 
r = (1 9 7 2 8)(3 6)(4 5) and 7r' = (19 7)(2 8)(3 6)(4 5) have e = 2. 
7r = (1297385)(46) and r' = (1297)(385)(46) have e = 3. 
7r= (197385 2)(46) and 7' = (1972)(385)(46) have e = 3. 
7r= (1948532)(67) and 7r' = (1932)(485)(67) have e = 4. 
7r = (1495832)(67) and 7r' = (14932)(58)(67) have e = 5. 
7r= (138697542) and r'= (138542)(697) have e = 6. 
Observe that every derangement r in Dn contains an extraction point unless r 
consists of a single cycle of the form 7 = (1 a2 Z), where Z is the ordered set 
{2, 3, ...., n - 1, n} - {a2}, written in decreasing order. For example, the 9-element 
derangement (1 5 9 8 7 6 4 3 2) has no extraction point. Since a2 can be any element 
of {2, 
... 
, n}, there are exactly n - 1 derangements of this type, all of which have 
sign (-1)n-1. We let X, denote the set of derangements of this form. Our problem 
reduces to finding a sign reversing involution f over T, = D, - X,. 
Suppose nr in Tn has extraction point e. Then the first cycle C1 of r ends with the 
(possibly empty) ordered subset Z consisting of the elements of {2, .. , e - 11 - {a2} 
written in decreasing order. Our sign reversing involution f : Tn -> Tn can then be 
succinctly described as follows: 
f (1 a2 X e Y Z)a (1 a2 X Z)(e Y)a, (3) 
where X and Y are ordered subsets, Y is nonempty, and a is the rest of the derange- 
ment r. 
Notice that since the number of cycles of 7r and f(7r) differ by one, they must 
be of opposite signs. The derangements on the left side of (3) are those for which the 
extraction point e is in the first cycle. In this case, Y must be nonempty, since otherwise 
"e Z" would be a longer decreasing sequence and e would not be the extraction point. 
The derangements on the right side of (3) are those for which the extraction point e is 
not in the first cycle (and must therefore be the leading element of the second cycle). 
In this case, Y is nonempty since iv is a derangement. Thus for any derangement i, 
the derangement f (i) is also written in standard form, with the same extraction point 
e and with the same associated ordered subset Z. Another way to see that n and f (i) 
have opposite signs is to notice that f(iv) = (xy)i (multiplying from left to right), 
where x is the last element of X (x = a2 when X is empty), and y is the last element 
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390 MATHEMATICS MAGAZINE 
of Y. Either way, f (f (r)) = 7r, and f is a well-defined, sign-reversing involution, as 
desired. I 
In summary, we have shown combinatorially that for all values of n, there are almost 
as many even derangements as odd derangements of n elements. Or to put it another 
way, when randomly choosing a derangement with at least five elements, the odds of 
having an even derangement are nearly even. 
Acknowledgment. We are indebted to Don Rawlings for bringing this problem to our attention and we thank 
Magnhild Lien, Will Murray, and the referees for many helpful ideas. 
REFERENCES 
1. R. A. Brualdi, Introductory Combinatorics, 3rd ed., Prentice-Hall, New Jersey, 1999. 
2. C. A. McCarthy and A. T. Benjamin, Determinants of the tournaments, this MAGAZINE, 69 (1996), 133-135. 
3. C. D. Olds, Odd and even derangements, Solution E907, Amer Math. Monthly, 57 (1950), 687-688. 
Volumes of Generalized Unit Balls 
XIANFU WANG 
UBC Okanagan 
3333 University Way, Kelowna 
B.C., Canada, V1V 1V7 
shawn.wang@ubc.ca 
Diamonds, cylinders, squares, stars, and balls. These geometric figures are familiar to 
undergraduate students, but what could they possibly have in common? One answer is: 
They are generalized balls. The standard Euclidean ball can be distorted into a variety 
of strange-shaped balls by linear and nonlinear transformations. The purpose of this 
note is to give a unified formula for computing the volumes of generalized unit balls 
in n-dimensional spaces. 
A generalized unit ball in Rn is described by the set 
BPI (1) 
where pl > 0, P2 > 0,..., Pn > 0. 
When the numbers pi1 
...., pn are all greater than or equal to 1, the unit ball B 
is convex. Since IxlP is not concave on [-1, 1] for 0 < p < 1, BP,...Pn is not neces- 
sarily convex anymore when n > 1. When p1 = P2 = ...= Pn, = p > 1, we obtain 
the usual 1, ball. The 12 ball is denoted by B. By choosing different numbers pi, we 
can alter the appearance of the generalized balls greatly, as shown in FIGURE 1 with 
examples in R3 
Motivated by an article by Folland [5], I derived a unified formula for calculating 
the volume of these balls. Although the volume formulas for the standard Euclidean 
ball TB and simplex have been known for a long time [4, pp. 208, 220], the unified 
formula is (relatively) new. It is surprising that no matter how strange the balls look, 
the volume of any ball can be computed by a single formula, as follows: 
THEOREM. Assume pi, ..., pn > O. The volume of the unit ball Bp 
is equal to 
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Arthur T. Benjamin

Curtis T. Bennett

Florence Newberger

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Recounting the Odds of an Even Derangement

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