For what values of n is the graph $K_n$ Eulerian?

Question

The coefficient of $ x^{50} $ in $ (1 + x)^{101} (1 - x + x^2)^{100} $ is:

Answer 1

An Eulerian graph is one in which there is a closed trail that visits every edge exactly once. This concept can be used to determine when the complete graph, $K_n$, is Eulerian.

To determine when $K_n$ is Eulerian, we need to utilize Euler's theorem for Eulerian circuits. According to this theorem, a connected graph has an Eulerian circuit if and only if all vertices have an even degree. For a complete graph, $K_n$, each vertex has a degree of $n - 1$ because it is connected to the remaining $n - 1$ vertices.

Thus, for $K_n$ to be Eulerian, each vertex must have an even degree:

$n - 1$ is even, which implies $n$ must be odd, so that $n - 1$ is even.

Therefore, $K_n$ is Eulerian if $n$ is even.

Let's verify the given options:

Odd values of n: Rejected because $n - 1$ is even when $n$ is even.
Even values of n: Correct, since $n$ even makes $n - 1$ odd.
Both odd and even values of n: Rejected because odd $n$ results in $n - 1$ odd, which does not satisfy the Eulerian graph condition.
None of the above: Rejected as we've determined even values of $n$ satisfy the condition.

For what values of n is the graph $K_n$ Eulerian?

Show Hint

The Correct Option is B

Solution and Explanation

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