The expression $x_{1}: x_{2}: x_{3}: \ldots: x_{n}$ only has a definite meaning if we indicate, for example, by using parentheses, the order in which the divisions should be performed. Thus, the expression can be rewritten as

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

where $i_{1}, i_{2}, j_{2}, \ldots, j_{n-k}$ represent some permutation of the indices $1,2, \ldots, n$, with $i_{1}<i_{2}<\ldots<i_{k}$ and $j_{1}<j_{2}<\ldots<j_{n-k}$.

For example,

$$
\left(x_{1}:\left(x_{2}: x_{3}\right)\right): x_{4}=\frac{x_{1} x_{3}}{x_{2} x_{4}}, \quad\left(\left(x_{1}: x_{2}\right): x_{3}\right): x_{4}=\frac{x_{1}}{x_{2} x_{3} x_{4}}
$$

How many different fractions of this form can be obtained from the above expression by placing the parentheses in different ways?
I. In the given expression, there are $n-1$ division symbols. The result of each division, if it appears as the divisor or dividend of a further division, is enclosed in parentheses. The result of the last division performed does not need to be enclosed in unnecessary parentheses, so the number of pairs of parentheses (opening and closing parentheses) is $n-2$. Therefore, the smallest possible value of $n$ is 3.

For $n=3$, there are 2 possible orders:

$$
\begin{aligned}
& \left(x_{1}: x_{2}\right): x_{3}=x_{1} /\left(x_{2} x_{3}\right) \\
& x_{1}:\left(x_{2}: x_{3}\right)=\left(x_{1} x_{3}\right) x_{2}
\end{aligned}
$$

The order of divisions: 1,2,

The order of divisions: 2,1.

To briefly describe the order, we assigned the index of the $x$ immediately preceding the division symbol to the division symbol.

For $n=4$, there are three times as many, i.e., 6 possible orders, since taking any division as the last one, there are 2 possible orders for the remaining 2 divisions. However, there is no difference between the orders 1, 3, 2 and 3, 1, 2, since when forming the quotient of two quotients, the order of the divisions in the numerator and denominator is not constrained by the parentheses. Thus, the 3 pairs of parentheses can be placed in 5 different ways:
order:
$1,2,3: \quad\left(\left(x_{1}: x_{2}\right): x_{3}\right): x_{4}$
$2,3,1: \quad x_{1}:\left(\left(x_{2}: x_{3}\right): x_{4}\right)$
$2,1,3:\left(x_{1}:\left(x_{2}: x_{3}\right)\right): x_{4}$
$3,2,1: \quad x_{1}:\left(x_{2}:\left(x_{3}: x_{4}\right)\right)$

$1,3,2$ and

$3,1,2: \quad\left(x_{1}: x_{2}\right):\left(x_{3}: x_{4}\right)$

Their number is 5, and the different fractions are respectively

$$
\frac{x_{1}}{x_{2} x_{3} x_{4}}, \quad \frac{x_{1} x_{3}}{x_{2} x_{4}}, \quad \frac{x_{1} x_{4}}{x_{2} x_{3}}, \quad \frac{x_{1} x_{3} x_{4}}{x_{2}}
$$

Their number is only 4, because the order 3,2,1 gives the same fraction as the order 2,1,3.

For $n=5$, it will be useful to utilize the previous results. For now, we only select the last division to be performed, and all the different forms of the dividend and divisor are given by (2) or (3). Thus, there are 4 sub-cases:

$$
\begin{array}{ll}
A:\left(x_{1}: x_{2}: x_{3}: x_{4}\right): x_{5} ; & C:\left(x_{1}: x_{2}\right):\left(x_{3}: x_{4}: x_{5}\right) \\
B:\left(x_{1}: x_{2}: x_{3}\right):\left(x_{4}: x_{5}\right) ; & D: x_{1}:\left(x_{2}: x_{3}: x_{4}: x_{5}\right)
\end{array}
$$

In case $A$, the further placement of parentheses always results in one of the fractions (3), and the result is always obtained by multiplying the denominator by $x_{5}$; thus, we get all different fractions, and we omit writing them down. In case $B$, we need to multiply the expressions (2) by $x_{5} / x_{4}$, resulting in the first two fractions of (4). In case $C$, the dividend is always $x_{1} / x_{2}$, and the divisor can be one of two types, obtainable from (2) by increasing each index by 2: $x_{3} /\left(x_{4} x_{5}\right)$ or $\left(x_{3} x_{5}\right) / x_{4}$. Taking the first one results in the third fraction of (4), and the fraction obtained with the second divisor has already been obtained in case $A$. Finally, in case $D$, the divisors are obtained from (3) by increasing each index by 1, but only the first divisor results in a new fraction, which is the fourth fraction of (4). Thus, we have found 8 different fractions.

$$
\frac{x_{1} x_{5}}{x_{2} x_{3} x_{4}}, \quad \frac{x_{1} x_{3} x_{5}}{x_{2} x_{4}}, \quad \frac{x_{1} x_{4} x_{5}}{x_{2} x_{3}}, \quad \frac{x_{1} x_{3} x_{4} x_{5}}{x_{2}}
$$

The 4 fractions in (4) are obtained from the fractions in (3) by multiplying the numerator by $x_{5}$, or the denominator by $x_{5}$. This is how the transition from $n=4$ to $n=5$ is made. In the same way, the 2 fractions in (2) are obtained from $x_{1} / x_{2}$ by adding $x_{3}$ in two different ways, and then from (2) to (3) the 4 fractions are obtained.

Thus, by placing $x_{3}, x_{4}$, and $x_{5}$ in all possible ways in the numerator and the denominator, and as we have seen, the number of different fractions doubles with each addition. $x_{1}$, however, can only be in the numerator, as it can only be the dividend, because there is no division symbol before it, and $x_{2}$ can only be in the denominator, because it is either the divisor itself or the dividend of a divisor formed from a sequence of $x_{i}$ numbers.

II. We claim that from the expression $x_{1}: x_{2}: \ldots: x_{n}$, by placing the parentheses in different ways, as many different fractions can be formed as there are ways to place $x_{3}, x_{4}, \ldots, x_{n}$ in the numerator and the denominator (and as we have seen, in every fraction $x_{1}$ is in the numerator and $x_{2}$ is in the denominator). There are $2^{n-2}$ such arrangements, because the number of unassigned numbers is $n-2$. We only need to show that every such fraction can be obtained by placing the parentheses in a suitable way.

We assume that our statement is true if $n$ is replaced by some integer $k(\geq 2)$, or any of the numbers $2,3, \ldots, k-1$. (This, as we have seen, is true for $k=5$.) Replacing $n$ with $k+1$, $x_{k+1}$ is either a factor in the denominator or the numerator of the desired fraction. The fraction obtained by omitting $x_{k+1}$ has at least one suitable placement of the $k-2$ pairs of parentheses according to the assumption. The fractions in which $x_{k+1}$ is in the denominator are obtained by taking $x_{k+1}$ as the divisor of the last division to be performed, and the remaining fraction as the dividend, i.e., by placing the new opening parenthesis before $x_{1}$ and the closing parenthesis after $x_{k}$ (before or after any existing parentheses). Indeed, when dividing a fraction, we multiply the denominator by the divisor.

If, however, $x_{k+1}$ needs to be in the numerator of the desired fraction, we write $x_{k+1}$ as the divisor of the last division to be performed, i.e., we place a new opening parenthesis immediately after the division symbol of this division, and the closing parenthesis after $x_{k+1}$. Then the expression of the form $A / B$ becomes: $A /\left(B / x_{k+1}\right)=(A / B) \cdot x_{k+1}$, so the numerator of $A / B$ must still be multiplied by $x_{k+1}$.

With this, we have proven the statement.

Vesztergombi Katalin (Budapest, Fazekas M. Gyak. G.) Szántó Ottó (Pécs, Zipernovszky K. Gépip. T.)

Note. Extending the idea used for the case $n=5$, it can be shown that the number of different placements of the $n-2$ pairs of parentheses is:

$$
\frac{1}{n}\binom{2 n-2}{n-1}
$$

We recommend that interested readers read the note 1 attached to the solution of problem 1187. II.[^0]


[^0]:    ${ }^{1}$ K.M.L. 26 (1963) 125-127. o.