Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench/Solution/Proof 2

by : $168$

 * The Magisterial Bench

Proof
Let $X$ be the set of all permutations of the bench.

Let $E, S, W$ be the sets of all permutations where the two Englishmen, Scotsmen and Welshmen are forced to sit next to each other, respectively.

Then $\map {\complement_X} E, \map {\complement_X} S, \map {\complement_X} W$ are the sets of all permutations where the two Englishmen, Scotsmen and Welshmen do not sit next to each other, respectively.

The number of permutations the question asks for is:

$E \cap S \cap W$ contains permutations of the form:
 * $\paren {E \ E} \ \paren {S \ S} \ \paren {W \ W} \ F \ I \ S \ A$

which can be considered as a set of $7$ objects that can be permuted in $7!$ ways, multiplied by $2^3$ to account for the fact that either of the two within the pairs can be arranged in $2$ different ways.

Thus:
 * $\size {E \cap S \cap W} = 7! \times 2^3 = 40 \, 320$

$E \cap S$ contains permutations of the form:
 * $\paren {E \ E} \ \paren {S \ S} \ W \ W \ F \ I \ S \ A$

where no restrictions are placed on the Welshmen.

This gives:
 * $\size {E \cap S} = 8! \times 2^2 = 161 \, 280$

Similarly:
 * $\size {S \cap W} = \size {E \cap W} = 161 \, 280$

$E$ contains all permutations of the form:
 * $\paren {E \ E} \ S \ S \ W \ W \ F \ I \ S \ A$

where the only restriction is the Englishmen must sit next to each other.

This gives:
 * $\size E = 9! \times 2 = 725 \, 760$

Similarly:
 * $\size S = \size W = 725 \, 760$

Finally $\size X = 10! = 3 \, 628 \, 800$.

Therefore:

is the number of permutations required.