2020 AMC 8

Complete problem set with solutions and individual problem pages

Problem 21 Hard

A game board consists of 64 squares that alternate in color between black and white. The figure below shows square P in the bottom row and square Q in the top row. A marker is placed at P. A step consists of moving the marker onto one of the adjoining white squares in the row above. How many 7-step paths are there from P to Q? (The figure shows a sample path.)

  • A.

    28

  • B.

    30

  • C.

    32

  • D.

    33

  • E.

    35

Answer:A

Solution 1

Notice that, to step onto any particular white square, the marker must have come from one of the 1 or 2 white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from P to that square is the sum of the number of ways to move from P to each of the white squares immediately beneath it (also called the Waterfall Method). To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from P to that square, as already stated).

The answer is therefore \boxed{\textbf{(A) }28}.

 

Solution 2

Suppose we "extend" the chessboard infinitely with 2 additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.

The total number of paths from P to Q, including invalid paths which cross over the red line, is then the number of paths which make 4 steps up-and-right and 3 steps up-and-left, which is \binom{4+3}{3} = \binom{7}{3} = 35. We need to subtract the number of invalid paths, i.e. the number of paths that pass through X or Y. To get to X, the marker has to make 3 up-and-right steps, after which it can proceed to Q with 3 steps up-and-left and 1 step up-and-right. Thus, the number of paths from P to Q that pass through X is 1 \cdot \binom{3+1}{3} = 4. Similarly, the number of paths that pass through Y is \binom{4+1}{1}\cdot 1 = 5. However, we have now double-counted the invalid paths which pass through both X and Y; from the diagram, it is clear that there are only 2 of these (as the marker can get from X to Y by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is 4+5-2=7, and the number of valid paths from P to Q is 35-7 = \boxed{\textbf{(A) }28}.