Combinatorics: choose 5 out of 10 colored balls I usually don't have any problems thinking about combinatorics but this problems answer doesn't seem correct.
There are $5$ black balls, $1$ red, $1$ green, $1$ blue, $1$ yellow and $1$ white.
In how many ways can you pick $5$ balls?
The answer is $2^5$ but why?
I'm comparing it with a lock with $4$ integers, which has the number of combinations of $10^4$.
Anyone got any explanations?
 A: You can get $0$, $1$, $2$, $3$, $4$, or $5$ black balls.


*

*Clearly, there is just one way to get $0$ black balls: you just get all the colored balls.

*There are five ways of getting only $1$ black ball: you could leave out the red, or the green or the blue, or the yellow, or the white ball.

*There are $5!/2!3!$ ways of getting $2$ black balls. You can leave out any combination of 2 out of the 5 colored balls.

*There are $5!/n!(5-n)!$ ways of getting $n$ black balls. (Did you get why?)


In the end, summing all the ways you can get 5 balls:
$$\frac{5!}{0!5!} + \frac{5!}{1!4!} + \cdots + \frac{5!}{5!5!}=2^5$$
Since the sum of the $n$-th row of Pascal's triangle is $2^n$.
A: For the set of ways to pick balls, there is either a red or none; and if there are no red balls, then there must be a black ball. Similarly for all other non-black colours.
Consider the five pairs of balls: red and black, green and black, blue and black, yellow and black, and white and black.
The number of ways to pick $5$ balls is the same as the number of ways to pick one ball from each of the five pairs: $2^5$.
