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3 x 4 x 4 = 48. Sadly at least two crackers must be exact duplicates of ones already made.
BUT if you could handle the sad look on people's faces, you could consider a missing item as a possible combination. This gives 4 x 5 x 5 = 125 different crackers. Many crackers would be missing one item, some missing two, and one cracker with nothing in at all! (This solution by Eric Chen.)
4x5x5=100 not 125
Also rather than Many and Some why not say 40 crackers will be missing 1 item and 14 will be missing 2?
Thanks both of you, I have amended the puzzle (use refresh): Absolutely Christmas Crackers Puzzle - Solution
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman