Finding total ways of selecting items such that no two are consecutive - permutation

Question

There are n items in a line. We have to find the number of ways the items can be selected with the restriction that no two consecutive items can be selected.
I tried to do it with recurrence relation but not able to reach on any. Please help me to solve the problem.

Answer 1

After searching on the net I got the solution of above problem.
Say there are N items. If N is even we can select almost N/2 items such that no two are consecutive and if N is odd, we can select almost (N+1)/2 items. Let K is maximum items that can be selected.
We can select 1 to K items.
For selecting one item, we can select any item.
For selecting two items, we keep N-2 items in a sequence. The circles below represent the items in the sequence. And we have total of N-1 spaces beginning from left of first item to right of last item. The spaces are represented by '_' underscore. If we select two of any space, and replace them with item, then we will have N items and selected two items will not be consecutive as no two spaces are consecutive.
_ o _ o _ o _ o _ o _ o _ o _ o _ o _ o _
For selecting p items, we will keep N-p items in the sequence which will result in N-p+1 spaces. We can select any p spaces from these N-p+1 spaces.
So total possible ways will become
NC1 + N-1C2 + N-2C3 + ... + N-K+1CK which is sum of first N Fibonacci numbers (1,1,2,3,5,...).
Also sum of first N Fibonacci numbers is F(n+2) - 1

Answer 2

It seems hard to understand the way you explained why it is Fibonacci series.
I have an easier way of explaining the same as below.
Suppose we express the number of combinations for n items as T(n).
If we do not select the first item then the number of combinations is same as the number of combinations for the remaining n-1 items, which is T(n-1).
If we select the first item (we cannot select the second item as it is consecutive to the first position) then the number of combinations is same as the number of combinations for remaining n-2 items, which is T(n-2).
Therefore the below conclusion.
T(n) = T(n-1) + T(n-2).
T(1) = 2 (1. selected and 2. not selected)
T(2) = 3 (1. both not selected, 2. only first selected, 3. only second selected)
This is a Fibonacci series and can be computed in O(n) time complexity.

Answer 3

I think you can do this by building an array of length n with each place on the array representing the number of ways the items could be selected if that place was the first one that was selected. (Selecting from left to right.)
Psuedo code (untested):
int[] list = new int[n];
int total = 0;
for(int position = n-1; position >= 0; position--)
{
list[position] = 1;
for(int subPos = position + 2; subPos < n; subPos++)
{
list[position] += list[subPos];
}
total += list[position];
}
Explanation:
The value in list[i] when this has finished running represents the number of ways of picking items from the line with item i being the left most item that is picked.
Obviously there is only one way of picking items such that the right most item is the left most item that is picked. If n = 5, the pickings could be represented like this in that case: 00001
Similarly, for the second most right item there is only one way to pick items such that it is the left most item: 00010.
For the third most right item, there is 1 way to pick it where you only pick that item, then you must add on the number of ways of picking each of the items that might be picked second (this is what the second loop is for). So that item would have: 00100 and 00101.
Fourth most right item: 01000, 01010, 01001.
Fith most right item (first item on the left): 10000, 10100, 10101, 10010, 10001.
So the array for n=5 would end up with these values: {5,3,2,1,1}
And the total would then be: 5 + 3 + 2 + 1 + 1 = 12

Answer 4

Its a simple solution.
Say you need to select 3 numbers out of first 100 natural numbers, such that no two are consecutive.
Consider first 98 natural numbers, and randomly select 3 natural numbers (a, b, c) in 98C3 ways.
We know 0 < a, b, c < 99 and |a-b|, |b-c|, |a-c| >= 1 (since a, b, c are different ).
let A=a+0 ; B=b+1 ; C=c+2 ;
So we now know difference between any two of A, B and C is greater than 1 (i.e A, B and C cannot be consecutive numbers).
And 0 < A, B, C<101. A, B and C satisfy all conditions for the required question.
So solution is 98 C 3.
Generalizing → selecting p items from N, such that no two are consecutive is (N-p-1) C p.

Answer 5

Ans) (n+1-r)C r
Suppose that we have n items. And we want to select or choose r items making sure that no two objects are selected consecutively. We are going to represent the objects with '0'.
So now we have a sequence of 00000000.......0000 {upto n terms} Now when ever we choose an item from the 'i'th position i={1,......,n} , let us represent it as 1. So if we choose one item from position 2, the new sequence becomes 01000000....000 {upto n terms}. Now if we have to choose r elements, there will be r 1 s in the binary sequence that we are designing. but the interesting point is that between two successive 1 s , there can be any natural number of 0's. So if we place n -r (r 1's will also be there ! ) zeros together, then there are total n-r+1 gaps (left and right of each zeros). Then we will have to place the r 1 s in position. We can choose the position for r 1 s from amongst n-r+1 positions. This can be done in (n+1-r)C r.
Hope you understand.