# Logic behind combinations with repetition?

I've read the stars and bars analogy, but it really doesn't make sense to me. The way I see things, combination with repetition of say 5 of the same color balls in 2 different boxes WITH REPETITION is the same as the combination of 7 different color balls in 2 different boxes WITHOUT REPETITION, meaning

$\dbinom{5}{2}_{same}$ should be equal to $\dbinom{7}{2}_{different}$, making the answer be $\dbinom{7}{2}_{same}$.

But this is not true, because the combination of same is not $\dbinom{5}{2}_{same}$ but instead $\dbinom{5 + 2 - 1}{2}_{same} = \dbinom{6}{2}_{same}$

• You seem to be mixing up indices. There are $0$ ways to put $7$ balls into $2$ boxes without repetition (that is, without putting more than one ball in the same box). – Marc van Leeuwen Mar 24 '14 at 7:36

Stars and Bars says that the number of ways to put $5$ indistinguishable balls into $2$ distinguishable boxes is $\binom{5+2-1}{2-1}$, which is $6$.