THE ANTIQUARY'S CHAIN.
(
Combination and Group Problems)
An antiquary possessed a number of curious old links, which he took to a
blacksmith, and told him to join together to form one straight piece of
chain, with the sole condition that the two circular links were not to
be together. The following illustration shows the appearance of the
chain and the form of each link. Now, supposing the owner should
separate the links again, and then take them to another smith and repeat
his former instructions exactly, what are the chances against the links
being put together exactly as they were by the first man? Remember that
every successive link can be joined on to another in one of two ways,
just as you can put a ring on your finger in two ways, or link your
forefingers and thumbs in two ways.
Answer:
THE number of ways in which nine things may be arranged in a row without
any restrictions is 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362,880. But we
are told that the two circular rings must never be together; therefore
we must deduct the number of times that this would occur. The number is
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40,320 x 2 = 80,640, because if we
consider the two circular links to be inseparably joined together they
become as one link, and eight links are capable of 40,320 arrangements;
but as these two links may always be put on in the orders AB or BA, we
have to double this number, it being a question of arrangement and not
of design. The deduction required reduces our total to 282,240. Then one
of our links is of a peculiar form, like an 8. We have therefore the
option of joining on either one end or the other on every occasion, so
we must double the last result. This brings up our total to 564,480.
We now come to the point to which I directed the reader's
attention--that every link may be put on in one of two ways. If we join
the first finger and thumb of our left hand horizontally, and then link
the first finger and thumb of the right hand, we see that the right
thumb may be either above or below. But in the case of our chain we must
remember that although that 8-shaped link has two independent _ends_ it
is like every other link in having only two _sides_--that is, you cannot
turn over one end without turning the other at the same time.
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!
Informational
