Sources page biographical material

5.I.2. COLOURING CHESSBOARD WITH NO REPEATS IN A LINE
New section. I know there is a general result that an n x n board can be n coloured if n satisfies some condition like n  1 or 5 (mod 6), but I don't recall any other old examples of the problem.
Dudeney. Problem 50: A problem in mosaics. Tit Bits 32 (11 Sep 1897) 439 & 33 (2 Oct 1897) 3. An 8 x 8 board with two adjacent corners omitted can be 8 coloured with no two in a row, column or diagonal. = Anon. & Dudeney; A chat with the Puzzle King; The Captain 2 (Dec? 1899) 314-320; 2:6 (Mar 1900) 598-599 & 3:1 (Apr 1900) 89.

Dudeney. AM. 1917. Prob. 302: A problem in mosaics, pp. 90 & 215-216. The solution to the previous problem is given and then it is asked to relay the tiles so that the omitted squares are the (3,3) and (3,6) cells.

Hummerston. Fun, Mirth & Mystery. 1924. Q.E.D. -- The office boy problem, Puzzle no. 30, pp. 82 & 176. Wants to mark the cells of a 4 x 4 board with no two the same in any 'straight line ..., either horizontally, vertically, or diagonally.' His answer is: ABCD,  CDEA, EABC, BCDA, which has no two the same on any short diagonal. The problem uses coins of values: A, B, C, D, E = 12, 30, 120, 24, 6 and the object is to maximize the total value of the arrangement. In fact, there are only two ways to 5-colour the board and they are mirror images. Four colours are used three times and one is used four times -- setting the value 120 on the latter cells gives the maximum value of 696.
5.J. SQUARED SQUARES, ETC.
NOTE. Perfect means no two squares are the same size. Compound means there is a squared subrectangle. Simple means not compound.
Dudeney. Puzzling Times at Solvamhall Castle: Lady Isabel's casket. London Mag. 7 (No. 42) (Jan 1902) 584 & 8 (No. 43) (Feb 1902) 56. = CP, prob. 40, pp. 67 & 191 193. Square into 12 unequal squares and a rectangle.

Max Dehn. Über die Zerlegung von Rechtecken in Rechtecke. Math. Annalen 57 (1903) 314 332. Long and technical. No examples. Shows sides must be parallel and commensurable.

Loyd. The patch quilt puzzle. Cyclopedia, 1914, pp. 39 & 344. = MPSL1, prob. 76, pp. 73 & 147 148. c= SLAHP: Building a patchquilt, pp. 30 & 92. 13 x 13 into 11 squares, not simple nor perfect. (Gardner, in 536, says this appeared in Loyd's "Our Puzzle Magazine", issue 1 (1907), ??NYS.)

Loyd. The darktown patch quilt party. Cyclopedia, 1914, pp. 65 & 347. 12 x 12 into 11 squares, not simple nor perfect, in two ways.

P. J. Federico. Squaring rectangles and squares -- A historical review with annotated bibliography. In: Graph Theory and Related Topics; ed. by J. A. Bondy & U. S. R. Murty; Academic Press, NY, 1979, pp. 173 196. Pp. 189 190 give the background to Moroń's work. Moroń later found the first example of Sprague but did not publish it.

Z. Moroń. O rozkładach prostokątów na kwadraty (In Polish) (On the dissection of a rectangle into squares). Przegląd Matematyczno Fizyczny (Warsaw) 3 (1925) 152 153. Decomposes rectangles into 9 and 10 unequal squares. (Translation provided by A. Mąkowski, 1p. Translation also available from M. Goldberg, ??NYS.)

M. Kraitchik. La Mathématique des Jeux, 1930, op. cit. in 4.A.2, p. 272. Gives Loyd's "Patch quilt puzzle" solution and Lusin's opinion that there is no perfect solution.

A. Schoenflies. Einführung in der analytische Geometrie der Ebene und des Raumes. 2nd ed., revised and extended by M. Dehn, Springer, Berlin, 1931. Appendix VI: Ungelöste Probleme der Analytischen Geometrie, pp. 402 411. Same results as in Dehn's 1903 paper.

Michio Abe. On the problem to cover simply and without gap the inside of a square with a finite number of squares which are all different from one another (in Japanese). Proc. Phys. Math. Soc. Japan 4 (1931) 359 366. ??NYS

Michio Abe. Same title (in English). Ibid. (3) 14 (1932) 385 387. Gives 191 x 195 rectangle into 11 squares. Shows there are squared rectangles arbitrarily close to squares.

Alfred Stöhr. Über Zerlegung von Rechtecken in inkongruente Quadrate. Schr. Math. Inst. und Inst. angew. Math. Univ. Berlin 4:5 (1939), Teubner, Leipzig, pp. 119 140. ??NYR. (This was his dissertation at the Univ. of Berlin.)

S. Chowla. The division of a rectangle into unequal squares. Math. Student 7 (1939) 69 70. Reconstructs Moroń's 9 square decomposition.

Minutes of the 203rd Meeting of the Trinity Mathematical Society (Cambridge) (13 Mar 1939). Minute Books, vol. III, pp. 244 246. Minutes of A. Stone's lecture: "Squaring the Square". Announces Brooks's example with 39 elements, side 4639, but containing a perfect subrectangle.

Minutes of the 204th Meeting of the Trinity Mathematical Society (Cambridge) (24 Apr 1939). Minute Books, vol. III, p. 248. Announcement by C. A. B. Smith that Tutte had found a perfect squared square with no perfect subrectangle.

R. Sprague. Recreation in Mathematics. Op. cit. in 4.A.1. 1963. The expanded foreword of the English edition adds comments on Dudeney's "Lady Isabel's Casket", which led to the following paper.

R. Sprague. Beispiel einer Zerlegung des Quadrats in lauter verschiedene Quadrate. Math. Zeitschr. 45 (1939) 607 608. First perfect squared square -- 55 elements, side 4205.

R. Sprague. Zur Abschätzung der Mindestzahl inkongruenter Quadrate, die ein gegebenes Rechteck ausfüllen. Math. Zeitschrift 46 (1940) 460 471. Tutte's 1979 commentary says this shows every rectangle with commensurable sides can be dissected into unequal squares.

A. H. Stone, proposer; M. Goldberg & W. T. Tutte, solvers. Problem E401. AMM 47:1 (Jan 1940) 48 & AMM 47:8 (Oct 1940) 570 572. Perfect squared square -- 28 elements, side 1015.

R. L. Brooks, C. A. B. Smith, A. H. Stone & W. T. Tutte. The dissection of rectangles into squares. Duke Math. J. 7 (1940) 312 340. = Selected Papers of W. T. Tutte; Charles Babbage Research Institute, St. Pierre, Manitoba, 1979; pp. 10-38, with commentary by Tutte on pp. 1-9. Tutte's 1979 commentary says Smith was perplexed by the solution of Dudeney's "Lady Isabel's Casket" -- see also his 1958 article.

A. H. Stone, proposer; Michael Goldberg, solver. Problem E476. AMM 48 (1941) 405 ??NYS & 49 (1942) 198-199. An isosceles right triangle can be dissected into 6 similar figures, all of different sizes. Editorial notes say that Douglas and Starke found a different solution and that one can replace 6 by any larger number, but it is not known if 6 is the least such. Stone asks if there is any solution where the smaller triangles have no common sides.

M. Kraitchik. Mathematical Recreations, op. cit. in 4.A.2, 1943. P. 198. Shows the compound perfect squared square with 26 elements and side 608 from Brooks, et al.

C. J. Bouwkamp. On the construction of simple perfect squared squares. Konink. Neder. Akad. van Wetensch. Proc. 50 (1947) 72-78 = Indag. Math. 9 (1947) 57-63. This criticised the method of Brooks, Smith, Stone & Tutte, but was later retracted.

Brooks, Smith, Stone & Tutte. A simple perfect square. Konink. Neder. Akad. van Wetensch. Proc. 50 (1947) 1300 1301. = Selected Papers of W. T. Tutte; Charles Babbage Research Institute, St. Pierre, Manitoba, 1979; pp. 99-100, with commentary by Tutte on p. 98. Bouwkamp had published several notes and was unable to make the authors' 1940 method work. Here they clarify the situation and give an example. One writer said they give details of Sprague's first example, but the example is not described as being the same as in Sprague.

W. T. Tutte. The dissection of equilateral triangles into equilateral triangles. Proc. Camb. Phil Soc. 44 (1948) 464 482. = Selected Papers of W. T. Tutte; Charles Babbage Research Institute, St. Pierre, Manitoba, 1979; pp. 106-125, with commentary by Tutte on pp. 101-105.

T. H. Willcocks, proposer and solver. Problem 7795. Fairy Chess Review 7:1 (Aug 1948) 97 & 106 (misnumberings for 5 & 14). Refers to prob. 7523 -- ??NYS. Finds compound perfect squares of orders 27, 27, 28 and 24.

T. H. Willcocks. A note on some perfect squares. Canadian J. Math. 3 (1951) 304 308. Describes the result in Fairy Chess Review prob. 7795.

T. H. Willcocks. Fairy Chess Review (Feb & Jun 1951). Prob. 8972. ??NYS -- cited and described by G. P. Jelliss; Prob. 44 -- A double squaring, G&PJ 2 (No. 17) (Oct 1999) 318-319. Squares of edges 3, 5, 9, 11, 14, 19, 20, 24, 31, 33, 36, 39, 42 can be formed into a 75 x 112 rectangle in two different ways. {These are reproduced, without attribution, as Fig. 21, p. 33 of Joseph S. Madachy; Madachy's Mathematical Recreations; Dover, 1979 (this is a corrected reprint of Mathematics on Vacation, 1966, ??NYS). The 1979 ed. has an errata slip inserted for p. 33 as the description of Fig. 21 was omitted in the text, but the erratum doesn't cite a source for the result.} The G&PJ problem then poses a new problem from Willcocks involving 21 squares to be made into a rectangle in two different ways -- it is not clear if these have to be the same shape.

M. Goldberg. The squaring of developable surfaces. SM 18 (1952) 17 24. Squares cylinder, Möbius strip, cone.

W. T. Tutte. Squaring the square. Guest column for SA (Nov 1958). c= Gardner's 2nd Book, pp. 186 209. The latter = Selected Papers of W. T. Tutte; Charles Babbage Research Institute, St. Pierre, Manitoba, 1979; pp. 244-266, with a note by Tutte on p. 244, but the references have been omitted. Historical account -- cites Dudeney as the original inspiration of Smith.

R. L. Hutchings & J. D. Blake. Problems drive 1962. Eureka 25 (Oct 1962) 20-21 & 34-35. Prob. G. Assemble squares of sides 2, 5, 7, 9, 16, 25, 28, 33, 36 into a rectangle. The rectangle is 69 x 61 and is not either of Moroń's examples.

W. T. Tutte. The quest of the perfect square. AMM 72:2, part II (Feb 1965) 29-35. = Selected Papers of W. T. Tutte; Charles Babbage Research Institute, St. Pierre, Manitoba, 1979; pp. 432-438, with brief commentary by Tutte on p. 431. General survey, updating his 1958 survey.

Blanche Descartes [pseud. of Cedric A. B. Smith]. Division of a square into rectangles. Eureka 34 (1971) 31-35. Surveys some history and Stone's dissection of an isosceles right triangle into 6 others of different sizes (see above). Tutte has a dissection of an equilateral triangle into 15 equilateral triangles -- but some of the pieces must have the same area so we consider up and down pointing triangles as + and - areas and then all the areas are different. Author then considers dissecting a square into incongruent but equiareal rectangles. He finds it can be done in n pieces for any n  7.

A. J. W. Duijvestijn. Simple perfect squared square of lowest order. J. Combinatorial Thy. B 25 (1978) 240 243. Finds a perfect square of minimal order 21.

A. J. W. Duijvestin, P. J. Federico & P. Leeuw. Compound perfect squares. AMM 89 (1982) 15 32. Shows Willcocks' example has the smallest order for a compound perfect square and is the only example of its order, 24.

5.J.1. MRS PERKINS'S QUILT
This is the problem of cutting a square into smaller squares.
Loyd. Cyclopedia, 1914, pp. 248 & 372, 307 & 380. Cut 3 x 3 into 6 squares: 2 x 2 and 5  1 x 1.

Dudeney. AM. 1917. Prob. 173: Mrs Perkins's quilt, pp. 47 & 180. Same as Loyd's "Patch quilt puzzle" in 5.J.

Dudeney. PCP. 1932. Prob. 117: Square of Squares, pp. 53 & 148 149. = 536, prob. 343, pp. 120 & 324 325. c= "Mrs Perkins's quilt".

N. J. Fine & I. Niven, proposers; F. Herzog, solver. Problem E724 -- Admissible Numbers. AMM 53 (1946) 271 & 54 (1947) 41 42. Cubical version.

J. H. Conway. Mrs Perkins's quilt. Proc. Camb. Phil. Soc. 60 (1964) 363 368.

G. B. Trustrum. Mrs Perkins's quilt. Ibid. 61 (1965) 7 11.

Ripley's Puzzles and Games. 1966. Pp. 16-17, item 7. "Can you divide a square into 6 perfect squares?" Answer as in Loyd.

Nick Lord. Note 72.11: Subdividing hypercubes. MG 72 (No. 459) (Mar 1988) 47 48. Gives an upper bound for impossible numbers in d dimensions.

David Tall. To prove or not to prove. Mathematics Review 1:3 (Jan 1991) 29-32. Tall regularly uses the question as an exercise in problem solving. About ten years earlier, a 14 year old girl pointed out that the problem doesn't clearly rule out rejoining pieces. E.g. by cutting along the diagonals and rejoining, one can make two squares.
5.J.2. CUBING THE CUBE
S. Chowla. Problem 1779. Math. Student 7 (1939) 80. (Solution given in Brooks, et al., Duke Math. J., op. cit. in 5.J, section 10.4, but they give no reference to a solution in Math. Student.)
5.J.3. TILING A SQUARE OF SIDE 70 WITH SQUARES OF SIDES

1, 2, ..., 24
J. R. Bitner. Use of Macros in Backtrack Programming. M.Sc. Thesis, ref. UIUCDCS R 74 687, Univ. of Illinois, Urbana Champaign, 1974, ??NYS. Shows such a tiling is impossible.
5.K. DERANGEMENTS
Let D(n) = the number of derangements of n things, i.e. permutations leaving no point fixed.
Eberhard Knobloch. Euler and the history of a problem in probability theory. Gaņita-Bhāratī [NOTE: ņ denotes an n with an underdot] (Bull. Ind. Soc. Hist. Math.) 6 (1984) 1 12. Discusses the history, noting that many 19C authors were unaware of Euler's work. There is some ambiguity in his descriptions due to early confusion of n as the number of cards and n as the number of the card on which a match first occurs. Describes numerous others who worked on the problem up to about 1900: De Moivre, Waring, Lambert, Laplace, Cantor, etc.
Pierre Rémond de Montmort. Essai d'analyse sur les jeux de hazards. (1708); Seconde edition revue & augmentee de plusieurs lettres, (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)); 2nd issue, Jombert & Quillau, 1714. Problèmes divers sur le jeu du trieze, pp. 54 64. In the original game, one has a deck of 52 cards and counts 1, 2, ..., 13 as one turns over the cards. If a card of rank i occurs at the i-th count, then the player wins. In general, one simplifies by assuming there are n distinct cards numbered 1, ..., n and one counts 1, ..., n. One can ask for the probability of winning at some time and of winning at the k-th draw. In 1708, Montmort already gives tables of the number of permutations of n cards such that one wins on the k-th draw, for n = 1, ..., 6. He gives various recurrences and the series expression for the probability and (more or less) finds its limit. In the 2nd ed., he gives a proof of the series expression, due to Nicholas Bernoulli, and John Bernoulli says he has found it also. Nicholas' solution covers the general case with repeated cards. [See: F. N. David; Games, Gods and Gambling; Griffin, London, 1962, pp. 144 146 & 157.] (Comtet and David say it is in the 1708 ed. I have seen it on pp. 54-64 of an edition which is uncertain, but probably 1708, ??NX. Knobloch cites 1713, pp. 130-143, but adds that Montmort gave the results without proofs in the 1708 ed. and includes several letters from and to John I and Nicholas I Bernoulli in the 1713 ed., pp. 290-324, and mentions the problem in his Preface -- ??NYS.)

Abraham de Moivre. The Doctrine of Chances: or, A Method of Calculating the Probability of Events in Play. W. Pearson for the Author, London, 1718. Prob. XXV, pp. 59-63. (= 2nd ed, H. Woodfall for the Author, London, 1738. Prob. XXXIV, pp. 95-98.) States and demonstrates the formula for finding the probability of p items to be correct and q items to be incorrect out of n items. One of his examples is the probability of six items being deranged being 53/144.

L. Euler. Calcul de la probabilité dans le jeu de rencontre. Mémoires de l'Académie des Sciences de Berlin (7) (1751(1753)) 255 270. = Opera Omnia (1) 7 (1923) 11 25. Obtains the series for the probability and notes it approaches 1/e.

L. Euler. Fragmenta ex Adversariis Mathematicis Deprompta. MS of 1750 1755. Pp. 287 288: Problema de permutationibus. First published in Opera Omnia (1) 7 (1923) 542 545. Obtains alternating series for D(n).

Ozanam-Montucla. 1778. Prob. 5, 1778: 125-126; 1803: 123-124; 1814: 108-109; 1840: omitted. Describes Jeu du Treize, where a person takes a whole deck and turns up the cards, counting 1, 2, ..., 13 as he goes. He wins if a card of rank i appears at the i th count. Montucla's description is brief and indicates there are several variations of the game. Hutton gives a lengthier description of one version. Cites Montmort for the probability of winning as .632..

L. Euler. Solutio quaestionis curiosae ex doctrina combinationum. (Mem. Acad. Sci. St. Pétersbourg 3 (1809/10(1811)) 57 64.) = Opera Omnia (1) 7 (1923) 435 440. (This was presented to the Acad. on 18 Oct 1779.) Shows D(n) = (n 1) [D(n 1) + D(n 2)] and D(n) = nD(n 1) + ( 1)ⁿ.

Ball. MRE. 1st ed., 1892. Pp. 106-107: The mousetrap and Treize. In the first, one puts out n cards in a circle and counts out. If the count k occurs on the k-th card, the card is removed and one starts again. Says Cayley and Steen have studied this. It looks a bit like a derangement question.

Bill Severn. Packs of Fun. 101 Unusual Things to Do with Playing Cards and To Know about Them. David McKay, NY, 1967. P. 24: Games for One: Up and down. Using a deck of 52 cards, count through 1, 2, ..., 13 four times. You lose if a card of rank i appears when you count i, i.e. you win if the cards are a generalized derangement. Though a natural extension of the problem, I can't recall seeing it treated, perhaps because it seems to get very messy. However, a quick investigation reveals that the probability of such a generalized derangement should approach e^-4.

Brian R. Stonebridge. Derangements of a multiset. Bull. Inst. Math. Appl. 28:3 (Mar 1992) 47-49. Gets a reasonable extension to multisets, i.e. sets with repeated elements.
5.K.1. DERANGED BOXES OF A, B AND A & B
Three boxes contain A or B or A & B, but they have been shifted about so each is in one of the other boxes. You can look at one item from one box to determine what is in all of them. This is just added and is certainly older than the examples below.
Simon Dresner. Science World Book of Brain Teasers. 1962. Op. cit. in 5.B.1. Prob. 84: Marble garble, pp. 40 & 110. Black and white marbles.

Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 26: Mexican jumping beans, pp. 40-41 & 96. Red and black beans in matchboxes. The problem continues with a Bertrand box paradox -- see 8.H.1.

Doubleday - 3. 1972. Open the box, pp. 147-148. Black and white marbles.
5.K.2. OTHER LOGIC PUZZLES BASED ON DERANGEMENTS
These typically involve a butcher, a baker and a brewer whose surnames are Butcher, Baker and Brewer, but no one has the profession of his name. I generally only state the beginning of the problem.

New section -- there must be older examples. Gardner, in an article: My ten favorite brainteasers in Games (collected in Games Big Book of Games, 1984, pp. 130-131) says this is one of his favorite problems. ??locate

I now see these lead to Latin rectangles, cf Section 5.I.
R. Turner, proposer: The sons of the dons; Eureka 2 (May 1939) 9-10. K. Tweedie, solver: On the problem of the sons of the dons. Eureka 4 (May 1940) 21-23. Six dons, in analysis, geometry, algebra, dynamics, physics and astronomy, each have a son who studies one of these subjects, but none studies the subject of his father. Several further restrictions, e.g., there are no two students who each study the subject of the other's father.

M. Adams. Puzzle Book. 1939. Prob. B.91: Easter bonnet, pp. 80 & 107. Women named Green, Black, Brown and White with 4 colours of hats and 4 colours of dresses, but name, hat and dress are always distinct.

J. B. Parker. Round the table. Eureka 5 (Jan 1941) 20-21 & 6 (May 1941) 11. Seven men, whose names are colours, with ties, socks and cars, being coloured with three of the names of other men and all colours used for each item, sitting at a table with eight places.

Anonymous. The umbrella problem. Eureka 9 (Apr 1947) 22 & 10 (Mar 1948) 25. Six men 'of negligible honesty' each go away with another's umbrella.

Jonathan Always. Puzzles to Puzzle You. Tandem, London, 1965. No. 30: Something about ties, pp. 16 & 74-75. Black, Green and Brown are wearing ties, but none has the colour of his name, remarked the green tie wearer to Mr. Black.

David Singmaster. The deranged secretary. If a secretary puts n letters all in wrong envelopes, how many envelopes must one open before one knows what is each of the unopened envelopes?

Problem proposal and solution 71.B. MG 71 (No. 455) (Mar 1987) 65 & 71 (No. 457) (Oct 1987) 238-239.

Open question. The Weekend Telegraph (11 Jun 1988) XV & (18 Jun 1988) XV.

5.K.3. CAYLEY'S MOUSETRAP
This is a solitaire (= patience) game developed by Cayley, based on Treize. Take a deck of cards, numbered 1, 2, ..., n, and shuffle them. Count through them. If a card does not match its count, put it on bottom and continue. If it matches, set it aside and start counting again from  1. One wins if all cards are set aside. In this case, pick up the deck and start a new game.
T. W. O. Richards, proposer; Richard I. Hess, solver. Prob. 1828. CM 19 (1993) 78 & 20 (1994) 77-78. Asks whether there is any arrangement which allows three or more consecutive wins. No theoretical solution. Searching finds one solution for n = 6 and n = 8 and 8 solutions for n = 9.
5.L. MÉNAGE PROBLEM
How many ways can n couples be seated, alternating sexes, with no couples adjacent?
A. Cayley. On a problem of arrangements. Proc. Roy. Soc. Edin. 9 (1878) 338 342. Problem raised by Tait. Uses inclusion/exclusion to get a closed sum.

T. Muir. On Professor Tait's problem of arrangements. Ibid., 382 387. Uses determinants to get a simple n term recurrence.

A. Cayley. Note on Mr. Muir's solution of a problem of arrangement. Ibid., 388 391. Uses generating function to simplify to a usable form.

T. Muir. Additional note on a problem of arrangement. Ibid., 11 (1882) 187 190. Obtains Laisant's 2nd order and 4th order recurrences.

É. Lucas. Théorie des Nombres. Gauthier Villars, Paris, 1891; reprinted by Blanchard, Paris, 1958. Section 123, example II, p. 215 & Note III, pp. 491 495. Lucas appears not to have known of the work of Cayley and Muir. He describes Laisant's results. The 2nd order, non homogeneous recurrence, on pp. 494 495, is attributed to Moreau.

C. Laisant. Sur deux problèmes de permutations. Bull. Soc. Math. de France 19 (1890 91) 105 108. General approach to problems of restricted occupancy. His work yields a 2nd order non-homogeneous recurrence and homogeneous 3rd and 4th order recurrences. He cites Lucas, but says Moreau's work is unpublished.

H. M. Taylor. A problem on arrangements. Messenger of Math. 32 (1903) 60 63. Gets almost to Muir & Laisant's 4th order recurrence.

J. Touchard. Sur un problème de permutations. C. R. Acad. Sci. Paris 198 (1934) 631 633. Solution in terms of a complicated integral. States the explicit summation.

I. Kaplansky. Solution of the "problème des ménages". Bull. Amer. Math. Soc. 49 (1943) 784 785. Obtains the now usual explicit summation.

I. Kaplansky & J. Riordan. The problème de ménages. SM 12 (1946) 113 124. Gives the history and a uniform approach.

J. Touchard. Permutations discordant with two given permutations. SM 19 (1953) 109 119. Says he prepared a 65pp MS developing the results announced in 1934 and rediscovered in Kaplansky and in Kaplansky & Riordan. Proves Kaplansky's lemma on selections by finding the generating functions which involve Chebyshev polynomials. Obtains the explicit summation, as done by Kaplansky. Extends to more general problems.

M. Wyman & L. Moser. On the 'problème des ménages'. Canadian J. Math. 10 (1958) 468 480. Analytic study. Updates the history -- 26 references. Gives table of values for n = 0 (1) 65.

Jacques Dutka. On the 'Problème des ménages'. Math. Intell. 8:3 (1986) 18 25 & 33. Thorough survey & history -- 25 references.

Kenneth P. Bogart & Peter G. Doyle. Non sexist solution of the ménage problem. AMM 93 (1986) 514 518. 14 references.