The number of ways in which 9 identical balls can be placed in three identical boxes is

Options

A. 55
B. `(9!)/((3!)^(4))`
C. `(9!)/((3!^(3))`
D. `12`

Answer

To solve the problem of distributing 9 identical balls into 3 identical boxes, we need to find the different combinations of non-negative integers \(x_1, x_2, x_3\) such that:

\[ x_1 + x_2 + x_3 = 9 \]

where \(x_1, x_2, x_3\) represent the number of balls in each box.

Since the boxes are identical, we need to consider the different partitions of the number 9 into at most 3 parts. We will list all the possible combinations:

1. **(9, 0, 0)**: All balls in one box.
2. **(8, 1, 0)**: Eight balls in one box, one in another.
3. **(7, 2, 0)**: Seven balls in one box, two in another.
4. **(7, 1, 1)**: Seven balls in one box, one in each of the others.
5. **(6, 3, 0)**: Six balls in one box, three in another.
6. **(6, 2, 1)**: Six balls in one box, two in another, and one in the last.
7. **(5, 4, 0)**: Five balls in one box, four in another.
8. **(5, 3, 1)**: Five balls in one box, three in another, and one in the last.
9. **(5, 2, 2)**: Five balls in one box, and two in each of the others.
10. **(4, 4, 1)**: Four balls in two boxes, and one in the last.
11. **(4, 3, 2)**: Four balls in one box, three in another, and two in the last.
12. **(3, 3, 3)**: Three balls in each box.

Now, we can count these combinations:

- (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)

In total, we have **12 distinct ways** to distribute the 9 identical balls into 3 identical boxes.

Thus, the answer is **12**.