Library / /708 of 740

, 2016

Combinatorics Difficulty 5.8 AIME, harder Prove it United States

Problem:

Rachel has two indistinguishable tokens, and places them on the first and second square of a $1 \times 6$ grid of squares. She can move the pieces in two ways:
- If a token has a free square in front of it, then she can move this token one square to the right.
- If the square immediately to the right of a token is occupied by the other token, then she can "leapfrog" the first token; she moves the first token two squares to the right, over the other token, so that it is on the square immediately to the right of the other token.

If a token reaches the 6th square, then it cannot move forward any more, and Rachel must move the other one until it reaches the 5th square. How many different sequences of moves for the tokens can Rachel make so that the two tokens end up on the 5th square and the 6th square?

Solution

Solution:

We put a marker on $(i, j)$ when a token is on the $i$ th and $j$ th square and $i > j$ . When the token in front/behind moves one step forward to a blank square, move the marker rightward/upward one unit correspondingly. When a "leapfrog" happens, the marker moves from $(x-1, x)$ to $(x, x+1)$ . We can translate this movement into: 1. move the marker upward to $(x, x)$ ; 2. move the marker rightward to $(x, x+1)$ . Thus, we set up a lattice path way from $(2,1)$ to $(6,5)$ staying under $y = x$ . This is a bijection since every intersection of the path way and $y = x$ indicates a "leapfrog". According to the definition of Catalan Number, the answer is the number of such lattice path ways, which is $C_{5} = 42$ .

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