A fashion is assigned to chose three models from eight different models for a fashion show next week. How many combinations of the three models could be chosen for next week?
To solve this I did 8*7*6=336.
But I am not sure what to do please help and also explain because I am not that good in maths....
I appreciate your help
Let's consider a smaller example. You are to pick two models out of 3.
You say there are three ways to choose the first and two ways to choose the second, which equals 3 * 2 = 6 possible ways to choose. Let's write those possibilities down.
[1] Pick A first, then B.
[2] Pick A first, then C.
[3] Pick B first, then A.
[4] Pick B first, then C.
[5] Pick C first, then A.
[6] Pick C first, then B.
Notice that you have the same pair of models whether you did [1] or [3], or whether you did [2] or [5], or whether you did [4] or [6]. So there are only three, not six, distinct combinations of models.
If the order in which you choose makes no difference to the result, then the simple multiplication that you did overcounts the result, and an adjustment is required. This adjustment is so common that it has a special notation.
\(\displaystyle m!\ MEANS\ m * (m - 1) * ... 1,\)
Notice that your answer of \(\displaystyle 8 * 7 * 6 = \dfrac{8 * 7 * 6 * 5 * 4 * 3 * 2 * 1}{5 * 4 * 3 * 2 * 1} = \dfrac{8!}{5!} = \dfrac{8!}{(8 - 3)!}.\)
But your answer was an overcount and must be adjusted.
Now, in your problem, you had to choose three models. not two. How many ways can we shuffle the same three models around in terms of choosing which came first, which came second, and which came third. Well once we are thinking THAT way, the order of choosing does make a difference and the formula is obviously
\(\displaystyle 3 * 2 = 3 * 2 * 1 = 3!.\)
So we need to adjust your answer by dividing it by 3!.
\(\displaystyle \dfrac{8!}{5!} \div 3! = \dfrac{8!}{5! * 3!} = \dfrac{8 * 7 * 6}{3 * 2} = 56.\)
This adjustment process is so common it has a special symbol \(\displaystyle \dbinom{8}{3}.\)
More generally \(\displaystyle \dbinom{n}{k} = \dfrac{n!}{(n- k)! * k!}.\)
