Compute:

  1. \(C(6, 3)\)

On a general note, I find that the most memorable and intuitive formula to calculate the number of combinations for me is the multiplicative formula:

\[\binom{n}{k} = \frac{n^\underline{k}}{k!}\]

With this formula it is relatively straightfoward to calculate combinations with low \(k\) in memory, like this example:

\[\frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20\]

  1. \(C(7, 4)\)

\[\frac{7 \cdot 6 \cdot 5 \cdot 4}{4 \cdot 3 \cdot 2 \cdot 1} = 35\]

  1. \(C(5, 1)\)

\[\frac{5}{1} = 5\]

  1. \(C(2, 4)\)

\[0\]