I have a statement that says:
In how many ways, can I choose $2$ different colored balls, if I have $3$
red, $4$ blue and $7$ yellow balls?
So, the order does not matter, because choose a red ball and a blue ball is the same thing that choose a blue ball and red ball, and I need a subgroup of $2$ elements of $14$ elements in total, also the balls must be of different colors.
According to this, I will use a combination $\frac{n!}{(n – k)!k !}$, then I will replace:
$$= \frac{14!}{12! \cdot 2!} = \frac{14 \cdot 13}{2} = 91$$
But my problem, is that the correct result must be $61$, and I would like to know, where my logic failed and how should it be done.
Best Answer
With your formula you are calculating the number of pairs of balls that you can obtain, but you are also counting pairs with balls of the same color.
You just simply have to discard them:
In how many ways you can pick 2 red balls? With your formula you can see that the number is 3.
In how many ways you can pick 2 blue balls? With your formula you can see that the number is 6.
In how many ways you can pick 2 yellow balls? With your formula you can see that the number is 21.
So, 91-3-6-21 is 61.