Maths Olympiad Prep

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Combinatorics Difficulty 5.6 AIME, harder Prove it Japan

How many sextuples (a,b,c,d,e,f)(a, b, c, d, e, f) of positive integers are there for which conditions a>b>c>d>e>fa > b > c > d > e > f and a+f=b+e=c+d=22a+f = b+e = c+d = 22 are satisfied?

Solution

To each nn chosen from the set of integers 1,2,,101, 2, \ldots, 10, there is a pair (m,n)(m, n) of integers satisfying the conditions m>nm > n and m+n=22m + n = 22. Therefore, there are exactly 1010 such pairs (m,n)(m, n). Finding sextuples (a,b,c,d,e,f)(a, b, c, d, e, f) satisfying the conditions of this problem is equivalent to finding triplets of pairs (m,n)(m, n) from this set of 1010 pairs. In fact, if we order the 33 pairs chosen as (a,f),(b,e),(c,d)(a, f), (b, e), (c, d) starting with the smallest and increasing the value of nn (hence f<e<df < e < d), then associate this triplet with the sextuple (a,b,c,d,e,f)(a, b, c, d, e, f). We then see that it satisfies the condition a>b>c>d>e>fa > b > c > d > e > f of the problem. Conversely, any such sextuple arises from a triplet of pairs from the set of 1010 pairs considered above. Therefore, the number we seek is given by 10C3=12010C3 = 120.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.