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Japan Mathematical Olympiad Initial Round

Combinatorics Difficulty 5.9 AIME, harder Prove it Japan

Find the value of
∑17Ca⋅17Cb⋅17Cc, \sum 17C_a \cdot 17C_b \cdot 17C_c,
where the sum is taken over all triples (a,b,c)(a, b, c) of non-negative integers satisfying a+b+c=5a+b+c=5.

Solution

2349060 \boxed{2349060}
Consider a particle moving on the xyxy-plane according to the following rule: if the particle is at the point (n,m)(n, m) where nn and mm are integers, it moves in one step either to the point (n+1,m)(n+1, m) or to the point (n,m+1)(n, m+1). Consider the set Ω\Omega of all paths that a particle can take to start at the origin (0,0)(0, 0) and take steps, following the rule as above to reach the point (46,5)(46, 5). It is easy to see that there are exactly 51C551C_5 such paths. Among these paths, there are exactly 17C8⋅17C8⋅17C817C_8 \cdot 17C_8 \cdot 17C_8 (where a+b+c=5a+b+c=5) paths which cross the lines x+y=17x+y=17 and x+y=34x+y=34 at the points (17−a,a)(17-a, a) and (34−a−b,a+b)(34-a-b, a+b), respectively. Since all the paths belonging to the set Ω\Omega cross these two lines at these points for some choice of aa and bb with 0≤a,b,0 \le a, b,

and a+b≤5a+b \le 5, we see that the value of the sum we seek is exactly the same as the number of paths belonging to the set Ω\Omega, which is 51C5=234906051C_5 = 2349060.

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