Thread: Choosing Non-Consecutive Integers from a list of N Integers

Prove that the number of ways choosing k integers from the integers 1,2,3...n, such that no two of the chosen integers are consecutive is:
(n-k+1) C k [as in : (n-k+1)CHOOSE(k) ]
with a suitable adjustment when n - k +1 < k

Originally Posted by Niall2

Prove that the number of ways choosing k integers from the integers 1,2,3...n, such that no two of the chosen integers are consecutive is:

Note that ; from the set the subset
can be represented by the string 001010101. If fact, any subset of four such that no two are consecutive can be represented by a 9-bit string of four 1’s and five zeros in which there are no adjacent 1’s.

Thus, from the set we can use an n-bit string consisting of k 1’s and n-k 0’s, having no adjacent 1’ to represent a subset of k elements containing no consecutive integers.

Ah. I can see how you can use an n-bit string to represent the problem.

Does that mean is the number of ways to arrange 1's and 0's in an n-bit string so that no two 1's are consecutive?
And if so, how can I prove that is the case?

Originally Posted by Niall2

Ah. I can see how you can use an n-bit string to represent the problem. Does that mean is the number of ways to arrange 1's and 0's in an n-bit string so that no two 1's are consecutive? And if so, how can I prove that is the case?

EXACTLY!

Let’s take a particular case: 1111000000 (four ones and six zeros).
We are going to use the zeros to separate the ones from each other: _0_0_0_0_0_0_. We have seven spaces into which we can put four ones.
That can be done in ways.
Now you generalize.

Brilliant...
EDIT: I now see completely

Thanks a lot