Stars and bars approach.

In how many ways can $b$ identical blue balls and $r$ identical red balls be distributed in $n$ distinct boxes ?

Here the answer given is :-

$\frac {(n+b-1)! (n+r-1)!}{(n-1)! (n-1)! b! r!}$

I understand how this answer is derived using stars and bars approach where we find the number of ways for each color and then multiply them.

However, I am not able to understand why the below answer is wrong ?

$\frac {(n+ (b+r) - 1)!} {(n-1)! b! r!}$.

In my second approach, I initially have $b+r$ identical balls and then divide them into $n$ distinct boxes. Why is this approach wrong?

• Why did you put a $b! r!$ in the denominator of your answer? – Mauve Jan 5 '18 at 20:00
• @Useless Because all blue balls and red balls are identical. – Zephyr Jan 5 '18 at 20:01
• Ok... Also, why did you omit the $(b+r)!$ that is in the denominator of $\binom{n+(b+r)-1}{n-1}$? – Mauve Jan 5 '18 at 20:03
• @Useless Because blue and red balls are same among themselves (balls of same color are identical). Not all (b+r) balls are identical. – Zephyr Jan 5 '18 at 20:05

The way you are counting, permuting $n-1$ bars, $b$ blue and $r$ red balls, counts $$\color{#55F}{\star}|\color{#55F}{\star}\color{#C00}{\star}|\color{#C00}{\star}$$ as different from $$\color{#55F}{\star}|\color{#C00}{\star}\color{#55F}{\star}|\color{#C00}{\star}$$ whereas the correct answer does not. It counts them as $$\color{#55F}{\star}|\color{#55F}{\star}|\quad\text{and}\quad|\color{#C00}{\star}|\color{#C00}{\star}$$