Hello,

I was teaching the children in my class how to solve the following:

4 different chocolate types. How many combination using two different chocolate types?

We deduced that for four separate chocolate types there were six combinations, and that if there had been five separate chocs, then 10 combinations. We then tabulated as follows;

1 0

2 1

3 3

4 6

5 10

6 15

7 21

Then, one of my gifted and talented year 5 children mentioned that they had heard that all combination puzzles could be solved with an equation, and then proceeded to ask me for the formula to solve the above calculation.

During the break, we used both pencil/paper and a calculator and concluded the following;

(n/2)-0.5 x n where n equals number of choc types.

For 6 chocolate types 6!/(2!x 4!) or (6!/4!)/2 (and I'm sure other combinations)

They all work, but unfortunately, I don't really know which is 'correct'.

If time permits, can someone guide me on how to respond to my pupils question?