The alphabet consists of 24 letters, and to describe each permutation of these letters, $24 \mathrm{~cm}^{2}$ of area is needed. How many times the surface area of the Earth is required to describe all permutations? (The radius of the Earth is $r=6370 \mathrm{~km}$.)
The number of 24-element permutations is $P_{24}=24!$ To describe these, according to the problem, $24!24 \mathrm{~cm}^{2}=24 \cdot 24!10^{-10} \mathrm{~km}^{2}$ of area is required. The surface area of the Earth is $4 r^{2} \pi=4 \cdot 6370^{2} \pi=4 \cdot 10^{2} \cdot 637^{2} \pi \mathrm{km}^{2}$. Therefore, the desired ratio is

$$
x=\frac{24 \cdot 24!10^{-10}}{4 \cdot 10^{2} \cdot 637^{2} \cdot \pi}=\frac{6 \cdot 24!10^{-12}}{637^{2} \pi}
$$

We must try to calculate this number as accurately as our 4-digit table allows. Simplifying the fraction would just be a waste of time. (It would be an even greater waste of time to attempt the prime factorization of $24!$, because after simplifying the fraction, it would be advisable to return to partial products for the sake of reducing errors and saving time.)

First, we calculate $\lg 6 \cdot 24!$. To reduce errors, but also to save time, it is advisable to determine partial products without the table, but only to the extent that these partial products do not contain more than 4 significant digits. Thus, $6 \cdot 24!=8!\cdot(9 \cdot 10 \cdot 11 \cdot 12) \cdot(13 \cdot 14 \cdot 15 \cdot 16) \cdot(17 \cdot 18 \cdot 19) \cdot(20 \cdot 21 \cdot 22) \cdot(23 \cdot 24 \cdot 6)=40320 \cdot 11880 \cdot 43680 \cdot 5814 \cdot 9240 \cdot 3312$.

If we add the logarithms of these 6 factors, we get that $\lg 6 \cdot 24!=24.5709$. After this, we can calculate $x$. (Instead of $2 \lg 637$, the more accurate $\lg 405769 \approx \lg 405800$ is used).

![](https://cdn.mathpix.com/cropped/2024_05_02_7ecfc570a98d6d4c680cg-1.jpg?height=227&width=482&top_left_y=762&top_left_x=816)

Thus, to describe all the permutations in question, the area of the Earth's surface would need to be 2,921,000 times larger. (With a 6-digit table, $x=2920570$).

Pál Szekerka (Bp. VI. Kölcsey g. IV. o. t.)

Note: We must marvel at the fact that most of the solvers did not use a logarithm table, but performed time-consuming calculations and often obtained inaccurate results because they took $\pi$ to be 3.14.