In how many ways can 9 identical balls be distributed into 3 identical

9 balls can be partitioned in the following ways,

{(9,0,0) (8,1,0) (7,2,0) (7,1,1) (6,3,0) (6,2,1) (5,4,0) (5,3,1) (5,2,2) (4,3,2) (4,4,1) (3,3,3)}

=> 12 ways

The order doesn't really matter as the 3 boxes are identical

Answer. B

I am just learning combinations and this is one of my weakest topics.
In a previous post, I learnt that for identical objects, we can use the stars and bars method or the n+r-1(c)r-1 formula. For distinct objects, we should use (number of groups)^(no. of items to distribute).

Since this question mentioned identical balls, I proceeded to follow the stars and bars method and go the answer 55. This is not the correct answer as per OA.

Can someone help me understand why the stars and bars method is not applicable here ? Is there any reading material where I can find consolidated study material (different methods used for different class of question) on P&C ?

Hi shaurya_gmat

In case where boxes were different, your method stands correct. However, in this problem along with balls the boxes are identical as well

For example, if the balls are distributed as (6,2,1) it won't make sense to count (1,2,6), (2,6,1), (2,1,6), (1,6,2), and (6,1,2) since the boxes are same. The stars and bars method stands valid if the distribution happened in distinct boxes.

Dear Krunaal why is it not 3^9? (3C1 ^9)

MartyMurray KarishmaB care to help?

That would be the answer if the balls and boxes were different; however, in this question, the balls and boxes are identical, so the number of cases will be much lower.

Care to help? Always! Realistically able to with time constraints of the day? Some times!
In any case, Bunuel has already provided you the explanation, so cheers!

3^9 would be the answer if placing a particular ball in a particular box would be different from placing that ball in another box.

In other words, if the boxes and balls were different from each other, we could answer the question in the following way.

We'd have 9 slots representing the 9 balls:

___ ___ ___ ___ ___ ___ ___ ___ ___

Then, for each ball, there would be three different boxes to put that ball in.

So, in that case, we would have the following:

_3_ × _3_ × _3_ × _3_ × _3_ × _3_ × _3_ × _3_ × _3_ = 3^9 ways of distributing 9 different balls in three different boxes

Notice, however, that there are not 3 different boxes to put the balls in, because the boxes are the same. So, for instance, putting ball 1 in box 1 would be the same as putting ball 1 in box 2, or box 3.

Also, the balls are the same. So, for example, putting a particular 3 balls together in a box is the same as putting 3 other balls together in a box.

Thus, there are not 3^9 ways to distribute the balls.

So, how many different ways to distribute them are there?

In this scenario, the only differentiator between distributions is the numbers of balls in boxes.

For example, 3 balls in each box is different from 4 in one box, 3 in another box, and 2 in another box.

So, the correct answer is 12 because there are 12 different ways for 3 integers to add up to 9.

(9, 0, 0) (8, 1, 0) (7, 2, 0) (7, 1, 1) (6, 3, 0) (6, 2, 1) (5, 4, 0) (5, 3, 1) (5, 2, 2) (4, 4, 1) (4, 3, 2) (3, 3, 3)

Correct answer: B

Hi Marty, thank you for your reply. Is there a formulaic approach to getting the answer (12) instead of writing all the cases out?

Thank you