In how many ways can five identical balls be distributed in three

I did not know the formula, so I did it with a different approach. I don't know where I went wrong.

Arrangement 1: 1 box has all 5 balls. So, arranging "500" in different ways 3!/2! (since 0 is repeated) = 3

Arrangement 2: 1 box has 4 balls, 1 other box has 1. Third has 0. So, arranging "410" in different ways 3! = 6

Arrangement 3: 1 box has 3 balls, other 2 boxes have 1 each. So, arranging "311" in different ways 3!/2! (since 1 is repeated) = 3

Arrangement 4: 1 box has 2 balls, second box has 2 and third has 1. So, arranging "221" in different ways 3!/2! (since 2 is repeated) = 3


I got a total of 15. What combination did I miss?

Darn it. I realized as soon as I hit Submit. I missed 320 combination. Which gives me the missing 6.

Can you give more examples where this equation can be used?

Since all five answer choices are relatively small, we can probably list and count all of the possible outcomes in under 2 minutes.
Let's list the outcomes the following way: ABC such that:
The first number is the number of balls in box A
The second number is the number of balls in box B
The third number is the number of balls in box C

We get:
500
410
401
320
302
311
230
203
221
212
140
104
113
131
122
050
005
041
014
032
023
DONE!

There are 21 outcomes

Answer: D

Hi BrentGMATPrepNow,
Thanks for a great explanation.
Just a small typo in the list 132 instead of 131, please correct it.

Good catch, thanks!
I've edited my response.
Kudos for you!!

The following is called the SEPARATOR method.

Five identical balls are to be separated into -- at most -- 3 groupings.
Thus, we need five balls and two separators:
OO|OO|O

Every arrangement of the elements above represents one way to distribute the 5 balls among three boxes A, B and C:
OO|OO|O = A gets 2 balls, B gets 2 balls, C gets 1 ball.
OO||OOO = A gets 2 balls, B gets 0 balls, C gets 3 balls.
OOOOO|| = A gets all 5 balls.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 7 elements above (the 5 identical balls and the 2 identical separators).
The number of ways to arrange 7 elements = 7!.
But when an arrangement includes identical elements, we must divide by the number of ways each set of identical arrangements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Here, we must divide by 5! to account for the 5 identical balls and by 2! to account for the 2 identical separators:
\(\frac{7!}{5!2!} = 21\)

Asked: In how many ways can five identical balls be distributed in three different boxes?

BBBBB||

Number of ways = 7!/5!2! = 21


IMO D

total ways to distribute five identical balls be distributed in three different boxes

5+3-1 C 3-1
7c2 ; 21 ways
option D

There are 2 ways to solve this question.

1. If you know the formula of "Theory of Partitioning", n+r-1 C r-1, where n are the number of objects and r will be the number of partitions. Always remember, 'r' will always be one less than the number of divisions to be done.

2. For the ones who do not want to remember the formula : We know that there are 5 IDENTICAL objects which need to be partitioned in 3 boxes,i.e. 2 separations. Total objects= 7. Hence 7! / 5!2!

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Asked: In how many ways can five identical balls be distributed in three different boxes?

BBBBB||
There are 5 identical balls and 2 partitions
Number of ways = 7!/5!2! = 21

IMO D

GMATGuruNY

Thanks
but what if the question is " In how many ways can five identical balls be distributed in three identical boxes?"....will your method work here?

The separator method is appropriate only if the boxes are considered distinct.
If the 3 boxes are considered identical, the most straightforward approach is simply to list all the ways in which the 5 identical balls can be separated into at most 3 groupings:
5-0-0 --> 5 in one box, the other two boxes empty
4-1-0 --> 4 in one box, 1 in another box, one box empty
3-2-0 --> 3 in one box, 2 in another box, one box empty
3-1-1 --> 3 in one box, 1 in each of the other two boxes
2-2-1 --> 2 in one box, 2 in another box, 1 in the third box
Total ways = 5