We will, for convenience, assume that each link has a black side and a side painted white.  Now, if it were stipulated that (with the chain lying on the table, and every successive link falling over its predecessor in the same way, as in the diagram) only the white sides should be uppermost as in A, then the answer would be 564,480, as above—­ignoring for the present all reversals of the completed chain.  If, however, the first link were allowed to be placed either side up, then we could have either A or B, and the answer would be 2 x 564,480 = 1,128,960; if two links might be placed either way up, the answer would be 4 x 564,480; if three links, then 8 x 564,480, and so on.  Since, therefore, every link may be placed either side up, the number will be 564,480 multiplied by 2^9, or by 512.  This raises our total to 289,013,760.

But there is still one more point to be considered.  We have not yet allowed for the fact that with any given arrangement three of the other arrangements may be obtained by simply turning the chain over through its entire length and by reversing the ends.  Thus C is really the same as A, and if we turn this page upside down, then A and C give two other arrangements that are still really identical.  Thus to get the correct answer to the puzzle we must divide our last total by 4, when we find that there are just 72,253,440 different ways in which the smith might have put those links together.  In other words, if the nine links had originally formed a piece of chain, and it was known that the two circular links were separated, then it would be 72,253,439 chances to 1 that the smith would not have put the links together again precisely as they were arranged before!

283.—­THE FIFTEEN DOMINOES.

The reader may have noticed that at each end of the line I give is a four, so that, if we like, we can form a ring instead of a line.  It can easily be proved that this must always be so.  Every line arrangement will make a circular arrangement if we like to join the ends.  Now, curious as it may at first appear, the following diagram exactly represents the conditions when we leave the doubles out of the question and devote our attention to forming circular arrangements.  Each number, or half domino, is in line with every other number, so that if we start at any one of the five numbers and go over all the lines of the pentagon once and once only we shall come back to the starting place, and the order of our route will give us one of the circular arrangements for the ten dominoes.  Take your pencil and follow out the following route, starting at the 4:  41304210234.  You have been over all the lines once only, and by repeating all these figures in this way, 41—­13—­30—­04—­42—­21—­10—­02—­23—­34, you get an arrangement of the dominoes (without the doubles) which will be perfectly clear.  Take other routes and you will get other arrangements.  If, therefore, we can ascertain just how many of these circular routes are obtainable from the pentagon, then the rest is very easy.