There are $2 n$ guests at a dinner party. Supposing that eh master and mistress of the house have fixed seats opposite one another and that there are two specified guests who must not be placed next to one another, show that the number of ways in which the company can be placed is $(2 n - 2)! \times (4 n^{2} - 6 n + 4) ˙$

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There are

2 n

guests at a dinner party. Supposing that eh master and mistress of the house have fixed seats opposite one another and that there are two specified guests who must not be placed next to one another, show that the number of ways in which the company can be placed is

(2 n - 2)! \times (4 n^{2} - 6 n + 4) ˙

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Question Text

There are

2 n

(2 n - 2)! \times (4 n^{2} - 6 n + 4) ˙

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