In how many ways can we place the numbers from $1$ to $100$ in a $2\times 50$ rectangle (divided into $100$ unit squares) so that any two consecutive numbers are always placed in squares with a common side?
Problem
Source: Tournament of Towns,Spring 2002, Senior O Level, P4
Tags: geometry, rectangle, combinatorics proposed, combinatorics
14.07.2016 07:28
Let f(n) is number of ways by which we place the numbers from 1 to 2n in a 2×n rectangle and place 1 on top-left cell.We can easily get the recursion f(n+1)=f(n)+1.Since f(1)=1, We can get f(n)=n.On the other hand, if we place 1 on the cell whose column is k(1<k<n), the number of ways is f(k-1)+f(n-k)=n-1.Therefore, whole number of ways is 4n+(n-1)・(2n-4)=2n^2-2n+4.This problem case is n=50, so 2・50^2-2・50+4=4904.
14.07.2016 08:07
question : (a) What does recursion mean ? (b) What's a recursion relation ?
. Thanks , kk108 .
14.07.2016 09:01
Hello, kk108! I am not good at english.So, I cannot understand your questions. What's the 'trivial' mean? You said that this problem is trivial?
14.07.2016 09:12
@above , trivial means easy . What I meant was that , you were saying that you used something called recursion, to solve that problem; so my question was , what does recursion mean ?This word was used in my math book as 'recurrence relation '; that's why I am asking . Thanks , kk108.
14.07.2016 09:39
I used the word 'recursion' as recurrence relation. By the way, how about your solution for this problem?