Solution
Ahead of the Two
Show the problem again
How many permutations of 1, 2, 3, 4, 5, 6 (written in a row) have both 1 and 3 appearing somewhere to the left of 2? Give an integer.
Worked solution
The answer is 240. Only the relative order of 1, 2 and 3 matters, and by symmetry each of the 3! = 6 relative orders occurs in the same number of permutations, namely 720/6 = 120. The condition says 2 is the last of the three, which holds in 2 of the 6 relative orders (1, 3, 2 and 3, 1, 2). So the count is 720 × 2/6 = 240.
Source: Classic counting exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.