Mathematical_Recreations-Kraitchik-2e

Permutational Problems 217
the minority, since otherwise some wife would be in the presence
of other men without her husband. (2) If h is any number
between 1 and 2n, at some time there must be just h
persons and the boat on the far side of the river. For after
the first crossing the number of persons on the far side of
the river is never increased by more than one at a time.
Suppose n is even and that n + 1 persons and the boat are
on the far side of the river, n - 1 persons are on the near side.
Since both numbers are odd there must be more women than
men on one side, which is only possible if there are no men on
that side. Hence all the men and one woman are on the far
side. But on the last crossing either one or two men must
have been taken from the near side, where they were in the
minority, which is not permitted.
If n is odd, let there be n people on each side, with the boat
on the far side. As before, all the men must be on one side.
If the men are on the far side, none of them may take the
boat back to get the women. If the men are on the near side,
they cannot get across, since the first man or men to cross
will be in a minority.
Almost the same arguments show that a boat holding 3
persons will not serve to transport more than 5 couples. The
only problem left is to determine whether x = 3 will serve
for 5 couples. This is settled by showing that there is a solution
when n = 5 and x = 3: (1) 3 wives cross over, (2) 1 wife
returns, (3) 2 wives cross, (4) 1 wife returns, (5) the 3 wives
who have crossed are joined by their husbands, (6) a couple
returns, (7) the 3 husbands cross, (8) a wife returns, (9) 3
wives cross, (10) a wife returns, (11) the last 2 wives cross.
Thus we have the following results (N being the least number
of crossings required for the given values of n and x):
n 2 3 4 5 >5
x 2 2 3 3 4
N 5 11 9 11 2n - 3