Talk:1983 AIME Problems/Problem 7

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.