What you're overlooking is that 6P4 = 6C4*4!kris610 wrote: I don't think 6P4 is appropriate here even if the order is important. In that case, you first find the number of ways you can *select* 4 seats from the available 6 -- 6C4. Once you find that, you multiply that by 4! i.e. each of the selections can be arranged in 24 different ways -- 6C4*4!
6P4 = 6!/(6-4)!
6C4 = 6!/4!(6-4)!
6C4 * 4! = 6!4!/4!(6-4)! = 6!/(6-4)! = 6P4
In general, we can always say:
nPk = n!/(n-k)!
nCk = n!/k!(n-k)!
So,
nPk = nCk * k!
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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