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34 votes

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

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

the formula for number of ways in which “n” identical items can be shared among r people is

c(n+r-1 , r-1) = c(n+r-1, n+r-1-(r-1)) [as c(n,r)=c(n, n-r)]

=c(n+r-1, n)

here in question it was given b identical blue balls therefore in our formulae, n=b and and r=n

i.e, c(b+n-1, b)

same with n=r..

but can some one explain how we derived forumlae for c(n+r-1 , r-1)…..

c(n+r-1 , r-1) = c(n+r-1, n+r-1-(r-1)) [as c(n,r)=c(n, n-r)]

=c(n+r-1, n)

here in question it was given b identical blue balls therefore in our formulae, n=b and and r=n

i.e, c(b+n-1, b)

same with n=r..

but can some one explain how we derived forumlae for c(n+r-1 , r-1)…..

0

edited
Sep 9, 2021
by Nirmalya Pratap 1

I tried pick and move strategy i got option D as answer :(

edit

after reading this https://math.stackexchange.com/questions/2339270/in-how-many-ways-can-b-blue-balls-and-r-red-balls-be-distributed-in-n-distinct-b

i figure it out what mistake i was doing

takeaway is read all the comment in gateoverflow it will give you so many different way to think about a single problem.

2

55 votes

Best answer

$r$ 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**)**.**

0

@Shakir Ahmad no your understanding is wrong,

This is formula is applicable if you have r identical objects and n distinct boxes. Here the problem is divided into two parts, first dividing the indistinguishable blue balls then dividing the indistinguishable red balls in n boxes respectively

0

5 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.

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

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.