In how many ways 20 identical balls can be put into 5 distinct bins such that each bin contains at least 2 balls?

Question

Firstly, I have to find the total no of ways in which 20 identical balls can be put into 5 distinct bins. Then, I have to subtract the no of cases when no of balls in at least one bin is less than 2. But, I'm not getting how to find that no. If possible, plz suggest me various ways to approach and solve this kind of problems. Plz help me with the complete solution to this problem.

Answer 1

As the balls the identical, I think we can just start the process with each box having two balls in each box. The identical nature of the balls tells us it can be done in only 1 way. So we are left with the task of filling 10 identical balls in 5 boxes.

The answer 5^20 is not right for identical balls, think of a less severe situation with 4 balls and 2 boxes, the problem is essentially asking in how many ways can you put the 4 balls in two different containers. Let's say the balls are b1,b2,b3 and b4 and say one 1 way of filling it is,

b1,b2 goes to container A b3,b4 goes to container B

another way of filling it proposes

b1,b3 goes to container A b2,b4 goes to container B

But in our case b1=b2=b3=b4 thereby both the cases are equivalent. I'm fact, there are 4C2 equivalent ways of filling the container 2 balls at a time. Same goes to the case of the original problem of filling 5 container with 20 identical balls, so there will be less number of possibilities.

Answer 2

Firstly give two balls to each box then use stick method to assign the remaining balls. The answer is 11C4+2(11C3)+11C2+11C1=726 Ask me if you don't get this!

Answer 3

Since the balls are identical, what matters here is how many balls are placed in each bin, not which ball is placed in which bin.

We can meet the requirement that at least two balls are placed in each bin by placing two balls in each of the five bins. That leaves ten balls to distribute to five boxes without restriction. The number of ways we can do this is the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$ in the nonnegative integers. A particular solution corresponds to the placement of $5 - 1 = 4$ addition signs in a row of $10$ ones. For instance, $$1 + + 1 1 1 1 + 1 1 1 + 1 1$$ corresponds to the solution $x_1 = 1$, $x_2 = 0$, $x_3 = 4$, $x_4 = 3$, $x_5 = 2$. The number of such solutions is the number of ways we can place four addition signs in a row of ten ones, which is $$\binom{10 + 5 - 1}{5 - 1} = \binom{14}{4}$$ since we must choose which four of the fourteen positions required for ten ones and four addition signs will be filled with addition signs.