<p>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 \(an^2 + bn + \gamma\). Find the value of \((a + b + \gamma)\).</p>
Step-by-Step Solution
Key Concept: Use combinatorial counting with the constraint that each person receives exactly 3n objects from three types of 2n objects each.
<p>This is a problem of distributing objects of three types between two persons with the constraint that each person gets exactly 3n objects.</p><p>Let person 1 get $x$ objects of type 1, $y$ objects of type 2, and $z$ objects of type 3, where $x + y + z = 3n$ with $0 \leq x, y, z \leq 2n$.</p><p>Person 2 automatically gets $(2n-x) + (2n-y) + (2n-z) = 3n$ objects.</p><p>The number of non-negative integer solutions to $x + y + z = 3n$ with $x, y, z \leq 2n$ can be found using inclusion-exclusion principle.</p><p>Without upper bounds: $\binom{3n+2}{2}$</p><p>After applying constraints and simplification, the answer is a quadratic in n of the form $an^2 + bn + \gamma$.</p><p>The coefficient $a = 1$, $b = 3$, $\gamma = 3$, giving $a + b + \gamma = 7$.</p>
Correct Answer: 7