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

by : $168$

 * The Magisterial Bench

Solution
According to :


 * $443 \, 520$

but by the working below it comes to $1 \, 532 \, 260$.

Proof
Without imposing any conditions, we can arrange $10$ men in one line in a total of $10! = 3 \, 628 \, 800$ different ways.

It remains to count how many of these are barred.

Let us regard two of a nationality together as one item.

Then we can consider:
 * $\paren {E \ E} \ \paren {S \ S} \ \paren {W \ W} \ F \ I \ S \ A$

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.

Hence we remove all such permutations, which is $7! \times 2^3 = 40 \, 320$.

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

where the Welshmen are not together.

This gives $8! \times 2^2$, but we must deduct the above arrangements or we will count them twice.

This gives $8! \times 2^2 - 7! \times 2^3 = 120 \, 960$.

The same applies to permutations of the form $E \ E \ \paren {S \ S} \ \paren {W \ W} \ F \ I \ S \ A$ and $\paren {E \ E} \ S \ S \ \paren {W \ W} \ F \ I \ S \ A$.

Hence two more deductions of $120 \, 960$.

Similarly we eliminate all permutations of the forms $\paren {E \ E} \ S \ S \ W \ W \ F \ I \ S \ A$, $E \ E \ \paren {S \ S} \ W \ W \ F \ I \ S \ A$ and $E \ E \ S \ S \ \paren {W \ W} \ F \ I \ S \ A$.

By similar reasoning, there are $9! \times 2 - 8! \times 2^2 = 564 \, 480$ in each of $3$ cases.

Hence we have:
 * $3 \, 628 \, 800 - 40 \, 320 - 3 \times 120 \, 960 - 3 \times 564 \, 480 = 1 \, 532 \, 260$