Difference between revisions of "Talk:1983 AIME Problems/Problem 7"
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420/552 = 35/46, leading to the same solution as before. | 420/552 = 35/46, leading to the same solution as before. | ||
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| + | [Note that the number of permutations with all separations >= 1 is 21*20=420, a much cleaner complementary-counting strategy for this problem, and no harder to obtain than the >=2 portion of the discussion above. Either can be obtained using a tokens-and-dividers argument.] | ||
Revision as of 13:57, 16 November 2008
I'm not sure how the posted solution arrived at the right answer, other than by luck.
The number of permutations (assuming the first knight is fixed) for which all three knights are separated by at least two seats is 18*17, not 20*19.
The number of permutations with at least one single-seat separation is 114. 6 of these permutions have two single-seat separations. 2 of these 6 have the fist knight 'between' the other two; 4 do not. 108 have exactly one single-seat separation. 4*18 of these 108 include the first knight in the single-seat separation and 2*18 do not.
18*17 + 114 = 420
The total number of permutations is 24*23=552.
420/552 = 35/46, leading to the same solution as before.
[Note that the number of permutations with all separations >= 1 is 21*20=420, a much cleaner complementary-counting strategy for this problem, and no harder to obtain than the >=2 portion of the discussion above. Either can be obtained using a tokens-and-dividers argument.]
