A BANK HOLIDAY PUZZLE.
(
Unicursal and Route Problems)
Two friends were spending their bank holiday on a cycling trip. Stopping
for a rest at a village inn, they consulted a route map, which is
represented in our illustration in an exceedingly simplified form, for
the puzzle is interesting enough without all the original complexities.
They started from the town in the top left-hand corner marked A. It will
be seen that there are one hundred and twenty such towns, all connected
by straight roads. Now they discovered that there are exactly 1,365
different routes by which they may reach their destination, always
travelling either due south or due east. The puzzle is to discover which
town is their destination.
Of course, if you find that there are more than 1,365 different routes
to a town it cannot be the right one.
Answer:
The simplest way is to write in the number of routes to all the towns in
this manner. Put a 1 on all the towns in the top row and in the first
column. Then the number of routes to any town will be the sum of the
routes to the town immediately above and to the town immediately to the
left. Thus the routes in the second row will be 1, 2, 3, 4, 5, 6, etc.,
in the third row, 1, 3, 6, 10, 15, 21, etc.; and so on with the other
rows. It will then be seen that the only town to which there are exactly
1,365 different routes is the twelfth town in the fifth row--the one
immediately over the letter E. This town was therefore the cyclist's
destination.
The general formula for the number of routes from one corner to the
corner diagonally opposite on any such rectangular reticulated
arrangement, under the conditions as to direction, is (m+n)!/m!n!,
where m is the number of towns on one side, less one, and n the number
on the other side, less one. Our solution involves the case where
there are 12 towns by 5. Therefore m = 11 and n = 4. Then the formula
gives us the answer 1,365 as above.
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