To solve the problem of dividing 2n objects of three different types between two persons such that each person gets 3n objects, we can follow these steps:
### Step 1: Understanding the Distribution
We have:
- 2n objects of type A
- 2n objects of type B
- 2n objects of type C
We need to divide these objects between two persons (let's call them P1 and P2) such that each person gets a total of 3n objects.
### Step 2: Setting Up the Equation
Let:
- P1 takes \(X\) objects of type A,
- \(Y\) objects of type B,
- \(Z\) objects of type C.
Since P1 must receive a total of 3n objects, we have the equation:
\[ X + Y + Z = 3n \]
where \(0 \leq X \leq 2n\), \(0 \leq Y \leq 2n\), and \(0 \leq Z \leq 2n\).
### Step 3: Finding Non-negative Solutions
The number of non-negative integer solutions to the equation \(X + Y + Z = 3n\) can be found using the stars and bars combinatorial method. The formula for the number of solutions is given by:
\[
\binom{P + k - 1}{k - 1}
\]
where \(P\) is the total we want (3n) and \(k\) is the number of variables (3 in this case). Thus, we have:
\[
\text{Number of solutions} = \binom{3n + 3 - 1}{3 - 1} = \binom{3n + 2}{2}
\]
### Step 4: Accounting for Over-Selection
However, we need to ensure that none of \(X\), \(Y\), or \(Z\) exceeds 2n. We will use the principle of inclusion-exclusion to account for cases where one or more variables exceed 2n.
1. **Case 1**: Suppose \(X > 2n\). Let \(X' = X - 2n\) (where \(X' \geq 1\)). The new equation becomes:
\[
X' + Y + Z = n - 1
\]
The number of non-negative solutions is:
\[
\binom{n - 1 + 3 - 1}{3 - 1} = \binom{n + 1}{2}
\]
2. **Case 2**: Similarly, for \(Y > 2n\) and \(Z > 2n\), we will have the same count of solutions:
\[
\text{Total for one variable exceeding } 2n = 3 \times \binom{n + 1}{2}
\]
3. **Case 3**: If two variables exceed 2n, say \(X > 2n\) and \(Y > 2n\), we set:
\[
X' = X - 2n, \quad Y' = Y - 2n
\]
The new equation becomes:
\[
X' + Y' + Z = n - 2
\]
The number of solutions is:
\[
\binom{n - 2 + 3 - 1}{3 - 1} = \binom{n + 1}{2}
\]
Since there are \(\binom{3}{2} = 3\) ways to choose which two variables exceed 2n, the total for this case is:
\[
3 \times \binom{n + 1}{2}
\]
### Step 5: Applying Inclusion-Exclusion
Using inclusion-exclusion, we calculate the total number of valid distributions:
\[
\text{Total valid solutions} = \binom{3n + 2}{2} - 3 \times \binom{n + 1}{2} + 3 \times \binom{n - 1}{2}
\]
### Step 6: Simplifying the Expression
After simplification, we can express the total number of ways as:
\[
\alpha n^2 + \beta n + \gamma
\]
where we identify the coefficients \(\alpha\), \(\beta\), and \(\gamma\).
### Step 7: Finding \(\alpha + \beta + \gamma\)
From the calculations, we find:
- \(\alpha = 3\)
- \(\beta = 3\)
- \(\gamma = 1\)
Thus, the final answer is:
\[
\alpha + \beta + \gamma = 3 + 3 + 1 = 7
\]
