TITLE: Combinatorics: choose 5 out of 10 colored balls
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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: $2^5$.