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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. $\frac{(n + (b + r) - 1)!} {n!\,(b + r - 1)}$
in Combinatory by Boss (16.3k points)
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+3

indistinguishable objects into n distinguishable boxes: $n_1 + n_2+ ..n_n = r$

$=C(n+r-1,r)$

Distribution of red balls and blue balls is independent, hence we can apply product rule.

(A) is the correct option!

0
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????
0

n-1+b+r C b+r   is incorrect because we do not have b+r indistinguishable objects. Any of the 𝑏 blue socks is distinguishable from any of the r red socks.

And the given solution is correct, since with no limit on the sizes of the boxes, you can think of this problem as first putting in the red socks, and then putting in the blue socks. That is, the number of ways I can distribute the blue socks is not at all affected by the number of ways I can distribute the red socks, and for every different way I distribute the red socks I can still distribute all the blue socks in all the different number of ways as if I had just blue socks. So, we multiply the number of ways you can distribute the red socks with the number of ways you can distribute the blue socks. And for each of those two tasks, you do use the 'putting indistinguishable objects into distinguishable boxes' formula.

2 Answers

+45 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).

by (261 points)
edited by
0
y not c???
0
In c: bbrr is one combination.
But since there are multiple boxes: <b | brr>, <bb | r | r>, <bb | rr> etc...are different combinations.
+1
Here shouldn't r red balls in n distinct box be C(n+r-1,r-1) and for b balls in n distinct box C(n+b-1, b-1) ?

Isn't http://math.stackexchange.com/questions/280598/in-how-many-ways-can-4-unlabeled-blue-balls-and-2-unlabeled-red-balls-be-packed same problem?
+3

In this case, the representation of a tuple as a sequence of stars and bars, with the bars dividing the stars into bins, is unchanged. The weakened restriction of nonnegativity (instead of positivity) means that one may place multiple bars between two stars, as well as placing bars before the first star or after the last star. Thus, for example, when n = 7 and k = 5, the tuple (4, 0, 1, 2, 0) may be represented by the following diagram.

★ ★ ★ ★ | | | ★ ★ |

Fig. 3: four bars give rise to five bins containing 4, 0, 1, 2, and 0 objects

This establishes a one-to-one correspondence between tuples of the desired form and selections with the replacement of k − 1 gaps from the n + 1 available gaps, or equivalently (k − 1)-element multisets drawn from a set of size n + 1. By definition, such objects are counted by the multi choose number.\left(\!{\tbinom  {n+1}{k-1}}\!\right)

To see that these objects are also counted by the binomial coefficient,\tbinom{n + k - 1}{n} observe that the desired arrangements consist of n + k − 1objects (n stars and k − 1 bars). Choosing the positions for the stars leaves exactly k − 1 spots left for the k − 1 bars. That is, choosing the positions for the stars determines the entire arrangement, so the arrangement is determined with {\displaystyle \tbinom{n + k - 1}{n} selections. Note that {\displaystyle {\tbinom {n+k-1}{n}}={\tbinom {n+k-1}{k-1}}}, reflecting the fact that one could also have determined the arrangement by choosing the positions for the k − 1 bars.

0

Please explain why is it not : nCr+b  (b+r)!/(b!.r!)

Select b+r boxes. Put b blue balls and r red balls into these b+r boxes.

Here n-(b+r) boxes will be empty. 

This answer is not in the option, so please explain why this is incorrect.

+1

Please explain how to derive this result

r red balls can be distributed into n distinct boxes in C(n+r-1,r) = .(n+r-1)! / (n-1)! r!

+1
its a standard result of distributing identical objects into distinct groups.

Example- distribution of balls in bins, balls in baskets.

It is same as no of integral solution solutions and combination with repetitions.
+2

(n+(b+r)−1)! / [(n-1)! * (b+r)!] .

can it could be an answer?

+4 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.
by Active (1.5k points)
edited by
+2
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????
Answer:

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