I understand how the solution is achieved, apart from the last step within the solutions. How do you get from: ([n-1] [n/2-1])/(n n/2) to: 1/2 * n/n ?
Hello (n-1) choose (n/2 - 1) is (n-1)! / [(n/2 - 1)! * (n/2)!]. Let's call this A. (n) choose (n/2) is n! / [(n/2)! * (n/2)!]. Let's call this B. When we take the ratio of A / B, the (n-1)! / n! from the numerators of each simplifies to 1/n. One of the (n/2)! cancels from each of the denominators. Finally, the (n/2 - 1)! / (n/2)! simplifies to 1 / (n/2). So in the end we get [1 / n] / [1 / (n/2)] = n / n * 1/2 = 1/2. Hope this helps! Andy
