Problem
COM-B2-M09-P015 Choosing Without Neighbours
#15
★★★★☆ Level 4 of 5
Prove that the number of ways to choose \(k\) numbers from \(\{1,2,\ldots,n\}\) with no two chosen numbers consecutive is \(\binom{n-k+1}{k}\).
If the chosen numbers are \(i_1<\cdots
Let the chosen numbers be \(i_1<\cdots
Although the solution is bijective, it explains the coefficient in the independence polynomial of a path.