CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Consider an equilateral triangle of side length n, which is divided into unit triangles, as shown. Let f(n) be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) revisits a triangle. An example of one such path is illustrated below for n = 5. Determine the value of f(2009)
  1. 720
  2. 540
  3. 2009!
  4. 2009


Solution

The correct option is B 540

Define a last triangle of a row as the triangle in the row that the path visits last. From the last triangle in row x, the path must move down and then directly across to the last triangle in row x + 1. Therefore, there is exactly one path through any given set of last triangles. For 1 < m < n - 1, there are m possible last triangles for the mth row. The last triangle of the last row is always in the centre. Therefore, f(n) = (n – 1)! And f(2009) = 2008!. Hence option (e)

Note:- You can also solve this by the Unitary approach

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More



footer-image