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May 2nd, 2018, 12:24 AM   #1
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The question is “in how many possible ways can 8 identical balls be distributed to 3 distinct boxes so that every box contains at least one ball?”

There is a hint saying that the case can be considered as putting two bars on a row of eight balls, like o o l o o o l o o o

the answer is 21. My calculation is 9C2 - (1 + 7 x 2) = 21

Could anyone explain the reason why the hint states so?

Moreover, if 8 different balls are used, would the calculation become 3^8 - ( 3C1 x 1^8 + 3C2 x 2^8 ) ?
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May 2nd, 2018, 09:25 AM   #2
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https://www.quora.com/What-is-the-nu...ne-being-empty
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May 2nd, 2018, 12:59 PM   #3
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Since each box has a least 1 ball,
then there are 5 balls left to distribute:
0 0 5
0 1 4
0 2 3
0 3 2
0 4 1
0 5 0 : 6
1 0 4
1 1 3
1 2 2
1 3 1
1 4 0 : 5
2 0 3
2 1 2
2 2 1
2 3 0 : 4
3 0 2
3 1 1
3 2 0 : 3
4 0 1
4 1 0 : 2
5 0 0 : 1

Using n(n+1)/2 formula:
SUM 1 to 6 = 6*7/2 = 21

Just aNUTter way of looking at it!
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May 3rd, 2018, 07:51 AM   #4
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Because the boxes are distinguishable, let's call them red, white, and blue.

$r = \text {# of balls in red box} \implies r \in \mathbb Z \text { and } 1 \le r \le 8 - 2 = 6.$

$w = \text {# of balls in white box} \implies w \in \mathbb Z \text { and } 1 \le w \le 8 - r - 1 = 7 - r.$

$b = \text {# of balls in blue box} \implies b \in \mathbb Z \text { and } b = 8 - r - w.$

In other words, b is fully determined by r and w.

Clearly, there are 6 possible values for r. Given any of one those values for r, there just as clearly are 7 - r possible values for w. And given any pair of values for r and w, there is only one possible value for b.

$\displaystyle \left ( \sum_{r=1}^6 w * b \right ) = \left ( \sum_{r =1} (7 - r) * 1 \right) = \left ( \sum_{r=1}^6 7 \right ) - \sum_{r = 1}^6 r =$

$\displaystyle 7 * \left ( \sum_{r=1}^6 1 \right ) - \dfrac{6 * 7}{2} = 7 * 6 - 21 = 42 - 21 = 21.$

The trick here is seeing that you must combine addition and multiplication.
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May 3rd, 2018, 08:37 AM   #5
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Quote:
Originally Posted by JeffM1 View Post
The trick here is seeing that you must combine addition and multiplication.
The real trick uses the diagram in the OP.

If you've got $N$ balls, there are $N-1$ "slots" between them. If you've got $K$ bags to fill with at least one ball, then you have $K-1$ "walls" to fill slots with to partition the identical balls into bags.

so there are $\dbinom{N-1}{K-1}$ ways to place these walls into slots and achieve at least one ball per bag.
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May 3rd, 2018, 08:40 AM   #6
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I like mine better: distribute 5 balls only...3 cheers for me!!
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May 3rd, 2018, 08:50 AM   #7
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Quote:
Originally Posted by Denis View Post
I like mine better: distribute 5 balls only...3 cheers for me!!
So if I try to apply this method to general $N$ balls and $K$ boxes I think i'd do

$d=N-K$

$s = \dfrac{(d+1)(d+2)}{2}$

In this problem we have $d=8-3=5$

$s = \dfrac{(6)(7)}{2} = 21$

If I have $4$ boxes however.

$d = 4$

$s = \dfrac{(5)(6)}{2}=15 \neq \dbinom{7}{3}=35$

So Brave Sir Denis how to generalize your method?
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May 3rd, 2018, 09:06 AM   #8
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Mine applies to the OP, not general cases...AHEM!
I respect your opinion, but if it ain't same as mine, then it's wrong
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