2018 AIME II Problem 8

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Concepts:partitions and compositionsmultiset permutationscasework

Difficulty rating: 2920

8.

A frog is positioned at the origin in the coordinate plane. From the point (x,y),(x, y), the frog can jump to any of the points (x+1,y),(x + 1, y), (x+2,y),(x + 2, y), (x,y+1),(x, y + 1), or (x,y+2).(x, y + 2). Find the number of distinct sequences of jumps in which the frog begins at (0,0)(0, 0) and ends at (4,4).(4, 4).

Solution:

The horizontal jumps are steps of 11 or 22 summing to 4,4, so as a multiset they are {1,1,1,1},\{1,1,1,1\}, {1,1,2},\{1,1,2\}, or {2,2},\{2,2\}, and the same holds for the vertical jumps. For any choice of the two multisets, every ordering of all the jumps is a valid sequence, and the number of orderings is the multinomial coefficient of the combined multiset.

The nine cases give (84)=70,7!4!2!=105 (twice),6!4!2!=15 (twice),\binom{8}{4} = 70, \quad \frac{7!}{4!\,2!} = 105 \text{ (twice)}, \quad \frac{6!}{4!\,2!} = 15 \text{ (twice)}, 6!2!2!=180,5!2!2!=30 (twice),(42)=6.\frac{6!}{2!\,2!} = 180, \quad \frac{5!}{2!\,2!} = 30 \text{ (twice)}, \quad \binom{4}{2} = 6.

The total is 70+2105+215+180+230+6=556.70 + 2 \cdot 105 + 2 \cdot 15 + 180 + 2 \cdot 30 + 6 = 556.

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