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 Probability and Statistics Basic Probability and Statistics Math Forum

 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
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Quote:
 Originally Posted by JeffM1 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. 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
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Quote:
 Originally Posted by Denis 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? 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  Tags counting Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post USAMO Reaper Probability and Statistics 2 February 9th, 2015 01:21 PM jbergin Advanced Statistics 1 December 28th, 2014 03:02 PM jbergin Probability and Statistics 1 December 23rd, 2014 07:58 AM panky Algebra 4 October 24th, 2012 10:02 AM soulsister5 Applied Math 5 May 10th, 2010 02:44 PM

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