52. In the expression $x_{1}: x_{2}: \ldots: x_{n}$ for indicating the order of operations, parentheses are placed and the result is written as a fraction:

$$
\frac{x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}}{x_{i_{1}} x_{l_{2}} \ldots x_{i_{n-k}}}
$$

(every letter $x_{1}, x_{2}, \ldots, x_{n}$ stands either in the numerator or in the denominator of the fraction). How many different expressions can be obtained in this way for all possible ways of placing parentheses?
52. Answer: $2^{n-2}$ fractions.

First of all, it is clear that in the resulting fraction, $x_{1}$ will be in the numerator. It is almost as obvious that $x_{2}$ will end up in the denominator in any arrangement of parentheses (the division sign in front of $x_{2}$ relates either to $x_{2}$ itself or to some expression containing $x_{2}$ in the numerator).

It turns out that the other letters $x_{3}, x_{4}, \ldots, x_{n}$ can be placed in the numerator or denominator in any way; hence, a total of $2^{n-2}$ fractions can be obtained: each of the $n-2$ letters $x_{3}, x_{4}, \ldots, x_{n}$ can independently be in the numerator or denominator.

We will prove this statement by induction. For $n=3$, we can obtain 2 fractions:

$$
\left(x_{1}: x_{2}\right): x_{3}=\frac{x_{1}}{x_{2} \cdot x_{3}} \quad \text { and } x_{1}:\left(x_{2}: x_{3}\right)=\frac{x_{1} x_{3}}{x_{2}}
$$

so the statement is true.

Assume that it is true for $n=k$ and prove it for $n=k+1$.

Let the expression $x_{1}: x_{2}: \ldots: x_{k}$ after some arrangement of parentheses be written as some fraction $A$. If in this expression we substitute $x_{k}: x_{k+1}$ for $x_{k}$, then $x_{k}$ will be in the same place as it was in the fraction $A$, and $x_{k+1}$ will be in the opposite place (if $x_{k}$ was in the denominator, then $x_{k+1}$ will be in the numerator and vice versa).

Now we will prove that we can add $x_{k+1}$ to the same place where $x_{k}$ stands. In the fraction $A$ after the arrangement of parentheses, there will necessarily be an expression of the form ( $p: x_{k}$ ), where $p$ is the letter $x_{k-1}$ or some parenthesis; by replacing ( $p: x_{k}$ ) with the expression $\left(\left(p: x_{k}\right): x_{k+1}\right)=p:\left(x_{k} \cdot x_{k+1}\right)$, we will obviously get the same fraction $A$, where $x_{k}$ is replaced by $x_{k} \cdot x_{k+1}$. Thus, the statement

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-124.jpg?height=372&width=396&top_left_y=1054&top_left_x=679)

Fig. 40 is proven.