If the integers from 1 to 10 are divided into 5 pairs, and the sums of the individual pairs are 11, 4, 7, 16, and 17, then the numbers in the individual pairs can be uniquely determined $\sqrt[1]{1}$. How many different groupings would result in the sums of the individual pairs being: 9, 10, 11, 12, 13? - Someone claims that the ten numbers can be arranged into five pairs in 945 different ways. Did they calculate correctly?[^0]


[^0]:    ${ }^{1}$ This problem was a puzzle in the 49th issue of Élet és Tudomány in 1965.
I. Solution. I. The representation of all numbers from 9 to 13 as the sum of two different numbers chosen from the integers 1 to 10:

$$
\begin{aligned}
& 9=1+8=2+7=3+6=4+5 \\
& 10=1+9=2+8=3+7=4+6 \\
& 11=1+10=2+9=3+8=4+7=5+6 \\
& 12=2+10=3+9=4+8=5+7 \\
& 13=3+10=4+9=5+8=6+7
\end{aligned}
$$

Representing 9 as $1+8$, 10 can be obtained from the unused numbers as $3+7$ and $4+6$. In the first case, 12 and 13 can only be given by $2+10$ and $4+9$, respectively, and the remaining $5+6$ forms 11, since the sum of the numbers 1 to 10 is equal to the sum of the numbers 9 to 13, so the sum of the two unused numbers in the first four representations always gives the remaining sum. In the second case, only the representations $11=2+9$ and $13=3+10$ are free for both terms.

Representing 9 as $2+7$, there are again two possibilities for 10: $1+9$ and $4+6$. In the first case, 12 can only be represented as $4+8$, and therefore 13 only as $3+10$, in the second case, $12=3+9$ is uniquely determined, and therefore $13=5+8$.

Representing 9 as $3+6$, either $10=1+9$ or $10=2+8$. In the first case, only $11=4+7$ and $13=5+8$ remain possible, in the second case, only $12=5+7$ and $13=4+9$. Finally, starting from the representation $9=4+5$, three representations remain for 10, but only two for 12, making it more efficient to continue with the pairing. Taking $12=2+10$, only $11=3+8$ and $13=6+7$ remain; if $12=3+9$, then $13=6+7,11=1+10$.

Thus, we have developed every representation of the number 9 into complete pairs in every possible way, generating all desired groupings, and we have obtained the following 8 groupings:

|  | 9 | 10 | 11 | 12 | 13 |
| ---: | ---: | ---: | ---: | ---: | ---: |
| I. | $1+8$ | $3+7$ | $5+6$ | $2+10$ | $4+9$ |
| II. | $1+8$ | $4+6$ | $2+9$ | $5+7$ | $3+10$ |
| III. | $2+7$ | $1+9$ | $5+6$ | $4+8$ | $3+10$ |
| IV. | $2+7$ | $4+6$ | $1+10$ | $3+9$ | $5+8$ |
| V. | $3+6$ | $1+9$ | $4+7$ | $2+10$ | $5+8$ |
| VII. | $3+6$ | $2+8$ | $1+10$ | $5+7$ | $4+9$ |
| VII. | $4+5$ | $1+9$ | $3+8$ | $2+10$ | $6+7$ |
| VIII. | $4+5$ | $2+8$ | $1+10$ | $3+9$ | $6+7$ |

Accordingly, the sums of the number pairs do not always uniquely determine the pairing of the numbers.

II. The pair for the number 1 can be chosen in 9 different ways from the other numbers. Each of these pairings can be continued in 7 different ways by choosing the smallest of the remaining numbers as the first element of the next pair, because in this way the second number of this pair can always be chosen in 7 different ways from the remaining 7 numbers, so the number of such further developments is $9 \cdot 7$. Similarly, by choosing the smallest of the unpaired numbers as the first element of the third and then the fourth pair, the second element of the third pair can be chosen in 5 different ways, and the second element of the fourth pair in 3 different ways, so the number of further developments first rises to $9 \cdot 7 \cdot 5$ and then to $9 \cdot 7 \cdot 5 \cdot 3=945$. This is the number of possibilities for the 5-pair arrangement, because we always considered every possibility for further development, these all differ from each other in at least one pair, and finally because the 4 pairs formed always determine the fifth pair as well. - According to this, the person calculated correctly.

Pál László (Pécs, Zipernovszky K. Mechanical Engineering High School I. class) Csur Endre (Budapest, Könyves Kálmán High School II. class)

Second solution to the second question of the problem. Two elements can be selected from $n$ different elements in $n(n-1) / 2$ ways, because after selecting the first element of the pair in $n$ ways, the second element can be chosen in $n-1$ ways, and the $n(n-1)$ pairings thus obtained give every pair twice, e.g., as $a, b$ and $b, a$. Therefore, if we perform the pairing of the 10 numbers written on the cards in such a way that we draw 2 cards at a time from a bag 4 times, each as the two elements of a pair, and also consider the remaining 2 numbers as a pair, this can be done in

$$
\frac{10 \cdot 9}{2} \cdot \frac{8 \cdot 7}{2} \cdot \frac{6 \cdot 5}{2} \cdot \frac{4 \cdot 3}{2}
$$

ways.

However, in this way we will get each of the 5-pair arrangements multiple times, specifically as many times as 5 different objects - here the pairs - can be arranged in a row. There are $2 \cdot 3 \cdot 4 \cdot 5$ possibilities for this, because by setting aside one object, the second can be placed in 2 ways relative to it: before it and after it, the third can be placed in 3 ways: before the first, before the second, and after the second, and so on. [^0]

Now we can get the number of ways to arrange the 10 numbers into 5 pairs - if we do not consider the order of the pairs - by dividing the above number by the latter:

$$
\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{2^{4} \cdot 2 \cdot 3 \cdot 4 \cdot 5}=5 \cdot 9 \cdot 7 \cdot 3=945
$$

Bárd Péter András (Budapest, Fazekas M. Grammar School I. class)


[^0]:    ${ }^{1}$ This problem was a puzzle in the 49th issue of Élet és Tudomány in 1966.