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

  1. $\frac{(n+b-1)!\,(n+r-1)!}{(n-1)!\,b!\,(n-1)!\,r!}$
  2. $\frac{(n+(b+r)-1)!}{(n-1)!\,(n-1)!\,(b+r)!}$
  3. $\frac{n!}{b!\,r!}$
  4. $\frac{(n + (b + r) - 1)!} {n!\,(b + r - 1)}$
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3 Answers

Best answer
57 votes
57 votes

$r$ red balls can be distributed into $n$-distinct boxes in $C(n+r-1,r) = \frac{\left(n+r-1\right)!}{\left(n-1\right)! r!}$

$b$ blue balls can be distributed in $C(n+b-1,b) = \frac{\left(n+b-1\right)!}{\left(n-1\right)! b!}$

By product rule total ways are  $\frac{\left(n+b-1\right)! \left(n+r-1\right)! }{\left(n-1\right)! b!\left(n-1\right)! r!}$

SO THE ANSWER IS  (A).

edited by
6 votes
6 votes
For $b$ Blue Balls:

$Box_1 + Box_2+Box_3+..........+Box_n =b$

its : $^{n+b-1}C_b$

For $r$ Red Balls:

$Box_1 + Box_2+Box_3+..........+Box_n =r$

its : $^{n+r-1}C_r$

Total Ways= $^{n+b-1}C_b$* $^{n+r-1}C_r$ (since blue and red balls are independent)

which is option A.
edited by
Answer:

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