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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?

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  • $\begingroup$ Why did you put a $b! r!$ in the denominator of your answer? $\endgroup$ – Mauve Jan 5 '18 at 20:00
  • $\begingroup$ @Useless Because all blue balls and red balls are identical. $\endgroup$ – Zephyr Jan 5 '18 at 20:01
  • $\begingroup$ Ok... Also, why did you omit the $(b+r)!$ that is in the denominator of $\binom{n+(b+r)-1}{n-1}$? $\endgroup$ – Mauve Jan 5 '18 at 20:03
  • $\begingroup$ @Useless Because blue and red balls are same among themselves (balls of same color are identical). Not all (b+r) balls are identical. $\endgroup$ – Zephyr Jan 5 '18 at 20:05
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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 $$ \color{#55F}{\star}|\color{#55F}{\star}|\quad\text{and}\quad|\color{#C00}{\star}|\color{#C00}{\star} $$

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  • $\begingroup$ I understand B means blue star and R means red star :P $\endgroup$ – Zephyr Jan 5 '18 at 20:28
  • $\begingroup$ @Zephyr: I just thought I'd use stars and bars $\endgroup$ – robjohn Jan 5 '18 at 20:33

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