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May 2nd, 2018, 12:24 AM  #1 
Member Joined: Nov 2012 Posts: 80 Thanks: 1  Counting
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 ) ? 
May 2nd, 2018, 09:25 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  
May 2nd, 2018, 12:59 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
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! 
May 3rd, 2018, 07:51 AM  #4 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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. 
May 3rd, 2018, 08:37 AM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  Quote:
If you've got $N$ balls, there are $N1$ "slots" between them. If you've got $K$ bags to fill with at least one ball, then you have $K1$ "walls" to fill slots with to partition the identical balls into bags. so there are $\dbinom{N1}{K1}$ ways to place these walls into slots and achieve at least one ball per bag.  
May 3rd, 2018, 08:40 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
I like mine better: distribute 5 balls only...3 cheers for me!!

May 3rd, 2018, 08:50 AM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  So if I try to apply this method to general $N$ balls and $K$ boxes I think i'd do $d=NK$ $s = \dfrac{(d+1)(d+2)}{2}$ In this problem we have $d=83=5$ $s = \dfrac{(6)(7)}{2} = 21$ 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? 
May 3rd, 2018, 09:06 AM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
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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