# Apologies for the simplicity of this question

• November 6th 2012, 12:33 PM
Connie
Apologies for the simplicity of this question
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?
• November 6th 2012, 12:47 PM
MarkFL
Re: Apologies for the simplicity of this question
I would tell them that with $n$ chocolate types chosen two at a time, there are $n$ choices for the first, $n-1$ choices for the second, and there are 2 ways to arrange them, so the total number of choices $N$ is:

$N=\frac{n(n-1)}{2}$

This may also be written using a binomial coefficient, or combination:

$N={n \choose 2}$

They may be interested to know that this is equivalent to:

$N=1+2+3+\cdots+(n-1)$
• November 6th 2012, 01:22 PM
Connie
Re: Apologies for the simplicity of this question
Sincere thanks for your prompt response.
• November 6th 2012, 01:30 PM
MarkFL
Re: Apologies for the simplicity of this question
Glad to assist, especially for someone who is obviously dedicated to instilling an enthusiasm for mathematics in young minds! Bravo!