maths riddle, page-22

You guys are incredible!
I'm a mathematician, and did some research on a generalisation of this problem a few years back - never completed it, but may try again one day.
The problem as stated (about the wives and carkeys etc) is a variation of the probleme des rencontres, posed by Pierre Remond de Montmort (1678-1719).
For this case, it could be restated as the number of different ways in which the wives go off with non-husbands - a huge number of permutations, the nearest integer to factorial 40 divided by e = 2.71828...
However, we want a probability. The total possible set of arrangements (permutations) is factorial 40 (40!); thus the difference, 40!-[[40!/e]] is the number of permutations in which at least one wife gets her husband. To make it a probability, divide back by 40!, and lo - we have the answer as stated above, 1 - 1/e which comes out to 0.6321 approx.
Hope this helps!
[And just because i'm a mathematician, don't trust my answer. I often get calculations muddled up!]

Thanks for brightening a slow day.