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

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)…..
edited

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

edit

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.

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

by

my doubt:   C(n-1+r,r)  this formula is applicable for the “distinct balls in the box problem” but in question it is not mentioned that the balls are distinct………

can anyone clarify my doubt plsss…………………….;;;;;;;;;;;;;

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

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.

In this type of questions, why are we not considering (b+r) balls into n distinct boxes ???

Instead of that we're doing r balls into n boxes * b balls into n boxes. What's the logic behind this doing so????
x1+x2+x3+.....Xn=b so c(n+b-1,b)

x1+x2+x3+.....xn=r so c(n+r-1,r)

so  multiplying both option a matches