The illustration shows clearly how the Weaver cut his square of beautiful cloth into four pieces of exactly the same size and shape, so that each piece contained an embroidered lion and castle unmutilated in any way. [Illustration] 15.--_The Cook's Puzzle._ There were four portions of warden pie and four portions of venison pasty to be distributed among eight out of eleven guests. But five out of the eleven will only eat the pie, four will only eat the pasty, and two are willing to eat of either. Any possible combination must fall into one of the following groups. (i.) Where the warden pie is distributed entirely among the five first mentioned; (ii.) where only one of the accommodating pair is given pie; (iii.) where the other of the pair is given pie; (iv.) where both of the pair are given pie. The numbers of combinations are: (i.) = 75, (ii.) = 50, (iii.) = 10, (iv.) = 10--making in all 145 ways of selecting the eight participants. A great many people will give the answer as 185, by overlooking the fact that in forty cases in class (iii.) precisely the same eight guests would be sharing the meal as in class (ii.), though the accommodating pair would be eating differently of the two dishes. This is the point that upset the calculations of the company. 16.--_The Sompnour's Puzzle._ The number that the Sompnour confided to the Wife of Bath was twenty-nine, and she was told to begin her count at the Doctor of Physic, who will be seen in the illustration standing the second on her right. The first count of twenty-nine falls on the Shipman, who steps out of the ring. The second count falls on the Doctor, who next steps out. The remaining three counts fall respectively on the Cook, the Sompnour, and the Miller. The ladies would, therefore, have been left in possession had it not been for the unfortunate error of the good Wife. Any multiple of 2,520 added to 29 would also have served the same purpose, beginning the count at the Doctor. 17.--_The Monk's Puzzle._ The Monk might have placed dogs in the kennels in two thousand nine hundred and twenty-six different ways, so that there should be ten dogs on every side. The number of dogs might vary from twenty to forty, and as long as the Monk kept his animals within these limits the thing was always possible. The general solution to this puzzle is difficult. I find that for _n_ dogs on every side of the square, the number of different ways is (_n_^4 + 10_n_^3 + 38_n_^2 + 62_n_ + 33) / 48, where _n_ is odd, and ((_n_^4 + 10_n_^3 + 38_n_^2 + 68_n_) / 48) + 1, where _n_ is even, if we count only those arrangements that are fundamentally different. But if we count all reversals and reflections as different, as the Monk himself did, then _n_ dogs (odd or even) may be placed in ((_n_^4 + 6_n_^3 + 14_n_^2 + 15_n_) / 6) + 1 ways. In order that there may be _n_ dogs on every side, the number must not be less than 2_n_ nor greater than 4_n_, but it may be any number within these limits.