Q.The number of ways of selecting 6 numbers from 1 to 49 such that their sum is even is

\dfrac{\binom{49}{6}}{2}

\binom{25}{6} + \binom{24}{3}\binom{25}{3}

2^{48}

\binom{49}{6} - \binom{25}{6}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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