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In how many ways can 4 red balls and 7 blue balls be arranged in 3 boxes where each box must contain at least 1 red ball and each box can contain less than or equal or 4 balls, under the following cases?

(1) Each ball is distinct and Each box is distinct

(2) Each ball is distinct and Each box is identical

My Attempt goes on like this

the only arrangement possible is
One box must contain 3 balls
and each of the remaining 2 boxes contain 4 balls

Options are

box          box        box
 1 red       1 red      2 red
 2 blue      3 blue     2 blue

OR

box          box        box
 2 red       1 red      1 red
 1 blue      3 blue     3 blue

Beyond this, I am not able to progress.

Can someone guide me how to approach these problems when (1) Each ball is distinct and Each box is distinct and (2) Each ball is distinct and Each box is identical.

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  • $\begingroup$ If each ball is distinct, then the colors add nothing to the question. $\endgroup$ – barak manos Feb 17 '16 at 8:42
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As noted, 1 box must contain 2 red balls. This box can contain 1,or 2 blue balls. If it contains 1 blue ball then each of the remaining boxes must contain 3 blue balls each. If it contains 2 blue balls, then one of the remaining boxes contains 2 blue balls and the final contains 3.

There are two cases: $(r^2b)(rb^3)(rb^3)$ or $(r^2b^2)(rb^2)(rb^3)$

(1) Each ball is distinct and Each box is distinct

We must count the ways to choose boxes and balls to go in to them. For the first case: $3$ choices of one box, $^4C_2$ choices of red balls for it, $^7C_1$ choices of blue, then by symmetry it doesn't matter which of the other box we next choose (we don't want to over count) and there are $^2C_1$ and $^6C_3$ ways to select red and blue balls to put into it; the remaining balls go in the last box. Thus the first case's count is: $3{^4C_2}{^7C_1}{^2C_1}{^6C_3}$

Can you count ways to select boxes and their balls in the second case? (Hint: not symmetric).

(2) Each ball is distinct and Each box is identical

Here we just count ways to select the balls.

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  • $\begingroup$ read your answer many times and practised with small examples to understand how the symmetry affect counting. $\endgroup$ – Kiran Feb 17 '16 at 10:25
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    $\begingroup$ Can you count ways to select boxes and their balls in the second case? it would be 3*4C2*7C2*2*2C1*5C2 , am i correct? $\endgroup$ – Kiran Feb 17 '16 at 10:26
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    $\begingroup$ Yes. You are. @Kirian $\endgroup$ – Graham Kemp Feb 17 '16 at 10:27
  • $\begingroup$ Thanks @Graham Kemp, so the total will be 3*4C2*7C1*2C1*6C3 + 3*4C2*7C2*2*2C1*5C2 which is 25200. Elizabeth's answer 18720 is wrong. Please confirm. Now I will try to understand your answer for the second part (Each ball is distinct and Each box is identical) and post it here $\endgroup$ – Kiran Feb 17 '16 at 10:31
  • $\begingroup$ can you also give more hint on how to count when Each ball is distinct and Each box is identical $\endgroup$ – Kiran Feb 17 '16 at 10:47
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Since balls are not identical we can label them as R1, R2, R3, R4, B1, B2, B3, B4, B5, B6, and B7.

Case 1: Boxes are identical

Let's first arrange the red balls into the boxes. Since there are 4 red balls and each box must contain at least one red ball, we must have a box with 2 red balls and 2 boxes with one red ball. The number of possible ways to do this is $\frac{4!}{2!} = 12$, because what we are doing is equivalent to choosing 2 red balls among {R1, R2, R3, R4} and putting them into the same box and the other two red balls will be put into the remaining boxes, one for each.

When arranging the blue balls there are basically two possibilities: i) either we will have 4 balls in the box that is containing 2 red balls, or ii) there are 3 balls in the box that is containing 2 red balls.

For i) there are $ 42 \frac{5!}{3! 2!} = 420 $ possible ways to do this. Because we can choose two blue balls to put into the box that is containing 2 red balls in 42 ways and the remaining 5 blue balls will be grouped into 3 blue balls and 2 blue balls.

For ii) there are $7 \frac{6!}{3! 3!} = 140$ possible ways to do this. Because we can choose the blue ball to put into the box that is containing 2 red balls in 7 different ways and the remaining 6 blue balls will be grouped into two groups of 3 blue balls.

Therefore, the total number of possible ways is 12x(140+120)=3120.

Case 2:Boxes are not identical.

We will group the balls in the same way we described above and there are 3 groups of balls in each time. Then we can put these groups of balls into Box1 Box2 and Box3 in 6 different ways hence the correct answer is 3120x6= 18720.

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  • $\begingroup$ I am clear with explanation given by @Graham Kemp . Trying to understand how your approach got different because this explanation also looks convincing. Can anyone help what has gone wrong here? $\endgroup$ – Kiran Feb 17 '16 at 12:29

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