Distributing Indistinguishable Balls

The problem asks for the number of ways to distribute 6 indistinguishable balls among 3 distinct people (A, B, and C) with the condition that each person receives at least one ball.

This is a classic problem involving distributing indistinguishable items into distinct bins with a minimum requirement.

Formulating the Equation

Let $x_A$, $x_B$, and $x_C$ be the number of balls received by person A, person B, and person C, respectively.

Since there are a total of 6 balls, the sum of the balls received by each person must equal 6:

$\qquad x_A + x_B + x_C = 6$

The condition that each person gets at least one ball translates to the following constraints:

• $x_A \ge 1$
• $x_B \ge 1$
• $x_C \ge 1$

Applying the Constraint (Stars and Bars Method)

To find the number of integer solutions that satisfy $x_A + x_B + x_C = 6$ with the constraints $x_A \ge 1$, $x_B \ge 1$, $x_C \ge 1$, we can use a transformation.

Let $y_A = x_A - 1$, $y_B = x_B - 1$, and $y_C = x_C - 1$.

Since $x_A \ge 1$, $x_B \ge 1$, and $x_C \ge 1$, the new variables $y_A$, $y_B$, and $y_C$ must be greater than or equal to 0:

• $y_A \ge 0$
• $y_B \ge 0$
• $y_C \ge 0$

Now, substitute $x_A = y_A + 1$, $x_B = y_B + 1$, and $x_C = y_C + 1$ into the original equation:

$\qquad (y_A + 1) + (y_B + 1) + (y_C + 1) = 6$

Simplify the equation:

$\qquad y_A + y_B + y_C + 3 = 6$

$\qquad y_A + y_B + y_C = 3$

The problem is now equivalent to finding the number of non-negative integer solutions to $y_A + y_B + y_C = 3$.

This can be solved using the stars and bars formula. The number of non-negative integer solutions to $x_1 + x_2 + \dots + x_k = n$ is given by $\binom{n+k-1}{k-1}$.

In our transformed equation $y_A + y_B + y_C = 3$, we have:

• $n = 3$ (the sum, representing the 'stars')
• $k = 3$ (the number of variables, representing the 'bins')

Using the stars and bars formula with $n=3$ and $k=3$:

Number of ways = $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2}$

Calculating the Combinations

Now, we calculate the value of $\binom{5}{2}$:

$\qquad \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)} = \frac{5 \times 4}{2 \times 1}$

$\qquad \binom{5}{2} = \frac{20}{2} = 10$

There are 10 non-negative integer solutions to $y_A + y_B + y_C = 3$, which means there are 10 ways to distribute the 6 indistinguishable balls among A, B, and C such that each person gets at least one ball.