Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench/Solution/Proof 1

by : $168$

 * The Magisterial Bench

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.


 * $(1): \quad$ We 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 $7! \times 2^3 = 40 \, 320$.


 * $(2): \quad$ 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 $8! \times 2^2 - 7! \times 2^3 = 120 \, 960$.


 * $(3): \quad$ The same applies to permutations of the form:
 * $E \ E \ \paren {S \ S} \ \paren {W \ W} \ F \ I \ S \ A$

This gives another deduction of $120 \, 960$.


 * $(4): \quad$ The same applies to permutations of the form:
 * $\paren {E \ E} \ S \ S \ \paren {W \ W} \ F \ I \ S \ A$.

Hence another deduction of $120 \, 960$.


 * $(5): \quad$ Similarly we eliminate all permutations of the form $\paren {E \ E} \ S \ S \ W \ W \ F \ I \ S \ A$.

By similar reasoning, there are $9! \times 2$ cases, from which we need to deduct the results of $(1)$, $(2)$ and $(3)$ (but not $(4)$).

This gives $9! \times 2 - 40 \, 320 - 2 \times 120 \, 960 = 443 \, 520$.


 * $(6): \quad$ Similarly we eliminate all permutations of the form $E \ E \ \paren {S \ S} \ W \ W \ F \ I \ S \ A$.

This gives another $443 \, 520$.


 * $(7): \quad$ Similarly we eliminate all permutations of the form $E \ E \ S \ S \ \paren {W \ W} \ F \ I \ S \ A$.

This gives another $443 \, 520$.

Add all these up together:


 * $40 \, 320 + 3 \times 120 \, 960 + 3 \times 443 \, 520 = 1 \, 733 \, 760$

Hence we have:
 * $3 \, 628 \, 800 - 1 \, 733 \, 760 = 1 \, 895 \, 040$