GATE IT 2008 | Question: 25

Question

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

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

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

C. $\dfrac{n!}{b!\,r!}$

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

Comments

$x_1+x_2+.....+x_n = r $ , gives $ ^{n+r-1}C_r$
$x_1+x_2+.....+x_n = b $ , gives $ ^{n+b-1}C_b$

It's not $^{n+b+r-1}C_{b+r}$ like I was thinking because
this will remove all permutations of red balls and green balls BUT
we need to remove permutations of red balls and green balls inside each box, not as a whole.

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Detailed Video Explanation with Complete Analysis: https://youtu.be/gYS1Ui0cJF0?t=4838&feature=shared

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Similar Question in GATE 1999: GATE 1999 Question

Similar Question from University of Toronto: University of Toronto Question

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First understand the difference between 

i) Choose r objects from n objects. C(n,r)

ii) Choose r objects from n kind of  objects. C(n-1+r,r)

 

Answer 1

$r$ 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).

Comments

Here we are considering that both colored balls can be in the same box at a time, right ?

And if we consider atmost 1 ball of any color can be in a box, then:

For r : $C(n-1+r,r)$

and for b : $C(n-r-1+b,b)$.

So,  $total = C(n-1+r,r) *C(n-r-1+b,b)$

Is this correct ?

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i didnt understand why we are considering same thing two times ?

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@pavansan where its two time? frst one is for red balls second one is for blue balls

Answer 2

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.

Comments

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????

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@MRINMOY_HALDER

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

you can refer this for your doubt...

Answer 3

Detailed Video Explanation with Complete Analysis: https://youtu.be/gYS1Ui0cJF0?t=4838&feature=shared

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Similar Question in GATE 1999: GATE 1999 Question

Similar Question from University of Toronto: University of Toronto Question

Answer 4

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

Comments

solve full pls

Answer 5

x= (distrubtion of b identical blue balls in n distinct bins ) = n-1+b C b

y= (distrubtion of r identical  red balls in n distinct bins ) = n-1+r C r 

total distribution=X AND Y

total distibution = x and y (task is complete when both x and y  happens) (note x and y are independednt they can occur independently 

total distribution = n-1+b C b  * n-1+r C r