Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**SPACKlick****Member**- Registered: 2013-08-21
- Posts: 3

There is an error on this page

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?

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,591

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

Offline