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You are given 20 identical balls and 5 bins that are coloured differently ( so that any two of the bins can be distinguished from one another). In how many ways can the balls be distributed into the bins in such a way that each bin has at least two balls?

My attempt: First of all , 2 balls are distributed in each bin. . Then I think that the remaining 10 balls can be distributed into either 1 bin or 2 bins or 3 bins and so on. Now if all 10 balls are distributed into 1 bins then there are 5 distint ways of doing so . If two bins are selected (10 ways) , then for each of this selection , the 10 balls can be distributed in the following way (9+1) , (8+2), (7+3) upto (5+5) and then permuting those two bins. Overall , my strategy is to decompose 10 as the sum of 1 , 2 , 3,.. 5 natural numbers in unique ways . Obviously the process is tedious , but doing this way my answer is 981 (the correct ans is 1001) . Is that calculation mistake ? or my method is wrong ? Please help

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  • $\begingroup$ I am new to this site and this is my first question . sorry for inconvenience (if any) as I don't know using latex for writing formulas . Any help or advice is appreciated $\endgroup$ Feb 6, 2020 at 12:06

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Your method is overly complicated. We can ignore $10$ of the balls as being mandated to appear in the $5$ bins. Then the problem reduces to the number of ways of placing $10$ balls in $5$ different bins without restrictions, which is by stars and bars $$\binom{10+5-1}{5-1}=1001$$

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Your approach looks solid.

As you say, putting two balls in each bin leaves $10$ unassigned balls and five bins for them to go in. By Stars and Bars there are $$\binom {14}{10}=1001$$ ways to do that.

I suspect you have an arithmetic error somewhere in your case by case analysis. Unfortunately, that way of doing things, while correct, can be quite error prone.

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Your strategy might work (at first sight I see no flaw) but as you said: the process is tedious.

Having left $10$ balls that must be divided among $5$ distinguishable bins comes to the same as finding the cardinality of: $$\{(a_1,a_2,a_3,a_4,a_5)\in\mathbb Z^5_{\geq0}\mid a_1+a_2+a_3+a_4+a_5=10\}$$ and there is a nice tool for that: stars and bars.

Have a look and give it a try.

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  • $\begingroup$ Ahh .. I didn't know that . Thank you all , very much. $\endgroup$ Feb 6, 2020 at 12:18
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It will be equal to solution of equation $$x_1+x_2+x_3+x_4+x_5=20 $$ Where $ x_i \ge 2$ let $x_i=y_i+2$ $$\therefore y_1+y_2+y_3+y_4+y_5=10$$ Number of solutions are ${10+5-1 \choose 10}$ For complete theory you can check at https://www.mathsdiscussion.com/distribution-of-identical-objects-into-distinct-groups/

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