Japan Mathematical Olympiad

To each $n$ chosen from the set of integers $1, 2, \ldots, 10$, there is a pair $(m, n)$ of integers satisfying the conditions $m > n$ and $m + n = 22$. Therefore, there are exactly $10$ such pairs $(m, n)$. Finding sextuples $(a, b, c, d, e, f)$ satisfying the conditions of this problem is equivalent to finding triplets of pairs $(m, n)$ from this set of $10$ pairs. In fact, if we order the $3$ pairs chosen as $(a, f), (b, e), (c, d)$ starting with the smallest and increasing the value of $n$ (hence $f < e < d$), then associate this triplet with the sextuple $(a, b, c, d, e, f)$. We then see that it satisfies the condition $a > b > c > d > e > f$ of the problem. Conversely, any such sextuple arises from a triplet of pairs from the set of $10$ pairs considered above. Therefore, the number we seek is given by $10C3 = 120$.