Spoiler Inside: Solution to Puzzle | SelectShow |
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These are fairly easily solved if the combinations that can lead to each scenario results are worked out and then the number of ways each such combination can occur.
There are 24 × 23 × 22 × 21 (255,024) possible selections. Percentages of this will be given to the nearest 0.01%. In scenario 1, it could be three solid chocolate and one of any remaining (including a fourth solid) or two solid chocolate and two honey. There are 6 ways to choose the first solid, 5 for the second, 4 for the third, and 21 ways for the other one. This, however, causes overlap of cases so it is better to split this into 6 for the first solid, 5 for the second, 4 for the third, and 18 ways for a non-solid last and 6, 5, 4, 3 ways for four solids. Respectively, the cases give 8,640 ways (since there are four ways to arrange each such combination: 6 × 5 × 4 × 18 × 4) and 360 ways (6 × 5 × 4 × 3). For two solid and two honey, it is 6 for the first solid, 5 for the second, 6 for the first honey, 5 for the second with 6 ways to arrange each of these giving 5400 (6 × 5 × 6 × 5 × 6). The grand total is 14,400 which is 5.65%. For scenario 2, it must be 2 nougat and 2 emetic. This is like above with the 2 solid and 2 honey. There are 5,400 ways (6 × 5 × 6 × 5 × 6) which is 2.12%. For scenario 3, it must be 3 emetic and anything but a solid with 6 possible arrangements of each combination. This gives 6 × 5 × 4 × 12 × 6 (5,760) ways with the fourth not being an emetic and 6 × 5 × 4 × 3 (360) if the fourth is an emetic. The total is 6,120 ways which is 2.40%. |