# on combinations

We've began a probability lesson in high school and talked about combinations, though there's one thing that's bothering me and my professor told me something along this line "it seems logical but it's not like that in math".

So let's say we have 4 white balls in a jar (so we're judging the balls through the colours), $\binom{4}{2}$ is the number of combinations possible when we pull out simultaneously $2$ white balls and I know it equals $6$ but in this case (in which the colour is the only difference).

Why? Shouldn't it be $1$? Isn't it always the same outcome? The balls are all white, white white = white white, no difference.

• It depends on the question. How many ways can we draw 2 unique white balls from the jar? ${4 \choose 2} = 6$. How many different color combinations can we draw? 1. – miradulo Mar 31 '18 at 6:19
• the unique term was needed, thanks – user531476 Mar 31 '18 at 6:24

The number of ways to pick two balls out of the jar is exactly $\binom{4}{2}=6$. And yes if you judge it by color, there is only one outcome, $\{w, w\}.$

Why it's $\binom{4}{2}$? Because when I'm choosing the balls, I don't even care whether they're identical or not. They're chosen as if I put them in a row and choose them according to their position numbers, which are unique. Clearly there are $\binom{4}{2}$ ways to choose a set of positions of size $2$.

Assume there are $10$ balls in the jar, $7$ white and $3$ red, and you choose two of them simultaneously. The balls are also secretly numbered from $1$ to $10$. It is then clear that there are ${10\choose 2}=45$ different ways to pick two balls, all of them equiprobable. But you don't know about the numbers and are only interested in the colors. You might ask "What is the probability that I pick two white balls", or "What is the probability that I pick two different balls?". When calculating these probabilities you have to go back to the $45$ equiprobable cases.

I your example there is just one color, hence there are no interesting probability questions to be asked. But nevertheless: There are ${4\choose2}=6$ equiprobable cases.

• To answer the questions, the numerators are $C(7,2)$ and $C(7,1)\cdot C(3,1)$ respectively. – linear_combinatori_probabi Mar 31 '18 at 13:48