It is obvious that there must be an odd number of crossings, and that if
the five husbands had not been jealous of one another the party might
have all got over in nine crossings. But no wife was to be in the
company of a man or men unless her husband was present. This entails two
more crossings, eleven in all.
The following shows how it might have been done. The capital letters
stand for the husbands, and the small letters for their respective
wives. The position of affairs is shown at the start, and after each
crossing between the left bank and the right, and the boat is
represented by the asterisk. So you can see at a glance that a, b, and c
went over at the first crossing, that b and c returned at the second
crossing, and so on.
ABCDE abcde *|..|
| |
1. ABCDE de |..|* abc
2. ABCDE bcde *|..| a
3. ABCDE e |..|* abcd
4. ABCDE de *|..| abc
5. DE de |,,|* ABC abc
6. CDE cde *|..| AB ab
7. cde |..|* ABCDE ab
8. bcde *|..| ABCDE a
9. e |..|* ABCDE abcd
10. bc e *|..| ABCDE a d
11. |..|* ABCDE abcde
There is a little subtlety concealed in the words "show the _quickest_
way."
Everybody correctly assumes that, as we are told nothing of the rowing
capabilities of the party, we must take it that they all row equally
well. But it is obvious that two such persons should row more quickly
than one.
Therefore in the second and third crossings two of the ladies should
take back the boat to fetch d, not one of them only. This does not
affect the number of landings, so no time is lost on that account. A
similar opportunity occurs in crossings 10 and 11, where the party again
had the option of sending over two ladies or one only.
To those who think they have solved the puzzle in nine crossings I would
say that in every case they will find that they are wrong. No such
jealous husband would, in the circumstances, send his wife over to the
other bank to a man or men, even if she assured him that she was coming
back next time in the boat. If readers will have this fact in mind, they
will at once discover their errors.