SOLUTION: Consider set of first 18 natural numbers. Five different numbers a1,a2,a3,a4,a5 are to be selected from set S under the condition that a(i +1) -a(i)≥2 for all i= 1,2,3,4.

This asks for the number of ways to choose 5 numbers from

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18

such that no two of the 5 are next to each other:

For illustration, suppose the 5 numbers chosen were
3,7,12,14,16.  Let's bracket those 5:

1,2,[3],4,5,6,[7],8,9,10,11,[12],13,[14],15,[16],17,18

For all but the last one, 16, let's bring the right-hand 
neighbor into the brackets:

1,2,[3,4],5,6,[7,8],9,10,11,[12,13],[14,15],[16],17,18

This is a sequence of 14 things, 5 brackets containing
numbers and 9 numbers not within brackets.  And 5 of the
14 indicate 5 choices according to the required condition
of no two chosen are next to each other.

Here is another example which shows why it is not necessary
to bring in a neighbor to the right of the largest number
chosen into a bracket.

Suppose the 5 numbers chosen were 1,5,10,12,18.  Let's bracket 
those 5:

[1],2,3,4,[5],6,7,8,9,[10],11,[12],13,14,15,16,17,[18]

For all but the last one, 18, let's bring the right-hand 
neighbor into the bracket:

[1,2],3,4,[5,6],7,8,9,[10,11],[12,13],14,15,16,17,[18]

This shows why we don't need to bring a right hand neighbor
into the last number chosen, because in this case there is
no right hand neighbor.  It is necessary that only the 
first four numbers chosen have a right hand neighbor, which
of course, must not have not been chosen as one of the 5. 

Notice that this is also a sequence of 14 "things", consisting
of 5 brackets containing numbers and 9 numbers not within 
brackets.  And 5 of these 14 indicate 5 choices according to 
the required condition that no two chosen are next to each other.

So the number of such choices is the number of ways to choose
5 positions from the 14 possible positions to place the 5 brackets.

Answer: 14C5 = 2002

Edwin