JCCDQBHWHCB011(i) Combinatorics : Uniform Over-counting (Division Principle)

     [ As an ice-breaker game, this is stupid!  Anyway … ]

Suggested Approach and Solution:-

     If the problem seems difficult, try a simpler problem first.  This is a useful heuristic / tactic, especially for permutation and combination problems.  Let’s say there are only six students.  Draw a diagram (a heuristic) with dots representing students, and arrows from each dot to other dots except the two dots beside it.  You will soon realize that each arrows goes in two directions.  This means there is over-counting by a factor of 2.  Whatever answer you get by counting the arrows forward must be divided by 2.

 

How many arrows are there?  (Heuristic: Think of an equivalent problem).  For each of the 6 dots, there are 6 – 3 = 3 arrows coming out (minus 3 because excluding itself and two beside it).

So the number of arrows is 6 x (6 – 3) = 18.  But we don’t really care about the direction of the arrows.  Because of the double-counting (and this is uniformly true for every pair of arrows), we must now divide this answer by 2 to get 9.  We can verify by counting from the diagram that this answer is correct.

Another approach: there are ⁶C₂ = 15 possible line segments (connections between two dots).  But we don’t count those on the perimeter of the hexagon, so we subtract 6 (because there are six sides all round the hexagon).  This gives 15 – 6 = 9.

For 17 students, the answer is: 17 x (17 – 3) / 2 = 119