The number of ways in which 2n objects of one type, 2n of another type and 2n of a third type can be divided between 2 persons so that each may have 3n objects is `alpha n^(2)+beta n +gamma`. Find the value of `(alpha+beta+gamma)`.

Answer

To solve the problem, we need to find the number of ways to divide 2n objects of three different types between two persons such that each person gets 3n objects. We will denote the three types of objects as A, B, and C, where each type has 2n objects.

### Step-by-Step Solution:

1. **Understanding the Total Distribution**:
We have a total of \(6n\) objects (2n of type A, 2n of type B, and 2n of type C). We need to divide these objects between two persons, say Person 1 and Person 2, such that each person receives exactly \(3n\) objects.

2. **Using the Stars and Bars Theorem**:
We can represent the distribution of objects using the stars and bars combinatorial method. We need to find the non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 = 3n \]
where \(x_1\), \(x_2\), and \(x_3\) represent the number of objects of type A, B, and C that Person 1 receives, respectively.

The number of non-negative integer solutions to this equation is given by:
\[ \binom{3n + 2}{2} \]

3. **Excluding Invalid Cases**:
However, we need to ensure that neither person receives more than 2n objects of any type. We need to exclude cases where one type exceeds 2n.

If Person 1 receives more than 2n objects of type A, we can set \(x_1' = x_1 - (2n + 1)\) (where \(x_1' \geq 0\)). The new equation becomes:
\[ x_1' + x_2 + x_3 = 3n - (2n + 1) = n - 1 \]
The number of solutions to this is:
\[ \binom{n + 1}{2} \]

Since there are three types of objects, we multiply this by 3:
\[ 3 \cdot \binom{n + 1}{2} \]

4. **Final Calculation**:
The total number of valid distributions is then:
\[ \binom{3n + 2}{2} - 3 \cdot \binom{n + 1}{2} \]

Expanding these binomial coefficients:
\[ \binom{3n + 2}{2} = \frac{(3n + 2)(3n + 1)}{2} \]
\[ \binom{n + 1}{2} = \frac{(n + 1)n}{2} \]

Therefore, the total becomes:
\[ \frac{(3n + 2)(3n + 1)}{2} - 3 \cdot \frac{(n + 1)n}{2} \]

Simplifying this:
\[ = \frac{(3n + 2)(3n + 1) - 3(n + 1)n}{2} \]
\[ = \frac{9n^2 + 5n + 2 - 3n^2 - 3n}{2} \]
\[ = \frac{6n^2 + 2n + 2}{2} = 3n^2 + n + 1 \]

5. **Identifying Coefficients**:
From the expression \(3n^2 + n + 1\), we can identify:
- \(\alpha = 3\)
- \(\beta = 1\)
- \(\gamma = 1\)

6. **Calculating \(\alpha + \beta + \gamma\)**:
Finally, we find:
\[ \alpha + \beta + \gamma = 3 + 1 + 1 = 5 \]

### Conclusion:
The value of \(\alpha + \beta + \gamma\) is **5**.