In how many ways can two different colored balls be chosen?

Question

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.

Answer 1

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.

Answer 2

think in this way.here i have total 14 balls in my box.so if i take any two of them then there will be 14C2 combinations.but it also includes the pair of same color.so,we must have to subtract them from them.the identical pair from red color is 3C2 ,from Blue is 4C2 , from yellow is 7C2. so the possible combination will 14C2-3C2-4C2-7C2=61.

Answer 3

Alternatively, the cases are: $$\underbrace{{3\choose 1}{4\choose 1}}_{1R1B}+\underbrace{{3\choose 1}{7\choose 1}}_{1R1Y}+\underbrace{{4\choose 1}{7\choose 1}}_{1B1Y}=61.$$

Answer 4

Another approach from my comment:

There are $7$ ways to choose a yellow ball and $7$ ways to choose the second ball, $4$ ways to choose a blue ball and $10$ ways to choose the second ball. $3$ ways to choose a red ball and $11$ ways to choose the second ball. This counts every choice twice.

Thus, we get $$ \frac{7\cdot7+4\cdot10+3\cdot11}2=61 $$