Problem Statement
How many non-negative odd integer solutions are there to the equation:
\[x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 96?\]The answer is in the form:
\[\binom{n}{k} \quad \text{where} \quad k < \frac{n}{2}\]Find \(nk\).
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Solution
This is essentially a stars and bars problem, but with a twist — the variables must be odd non-negative integers.
Step 1: Reduce the Problem
We know any odd non-negative integer can be written in the form:
\[x_i = 2k_i + 1 \quad \text{where } k_i \geq 0\]So, substituting this in for each \(x_i\):
\[x_1 + x_2 + \dots + x_6 = (2k_1 + 1) + (2k_2 + 1) + \dots + (2k_6 + 1) = 2(k_1 + k_2 + \dots + k_6) + 6\]Set:
\[2(k_1 + \dots + k_6) + 6 = 96 \Rightarrow k_1 + \dots + k_6 = \frac{96 - 6}{2} = 45\]So now, the problem reduces to:
In how many ways can we write 45 as the sum of 6 non-negative integers?
This is a classic stars and bars problem. The number of non-negative integer solutions to:
\[k_1 + \dots + k_6 = 45\]is:
\[\binom{45 + 6 - 1}{6 - 1} = \binom{50}{5}\]Step 2: Extracting \(n\) and \(k\)
We are told the answer is \(\binom{n}{k}\), where \(k < \frac{n}{2}\).
Here, \(\binom{50}{5}\) satisfies that, since \(5 < 25\).
So, \(n = 50\), \(k = 5\), and:
Final Answer: The number of required solutions are \(\boxed{250}\)
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