20 balls in 5 different bins, at least 2 per bin 
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
 A: 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
$$\binom{10+5-1}{5-1}=1001$$
A: 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.
A: 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.
A: 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/
