This proof uses permutations. Choose the first \(k\) elements of a random permutation. Let \(a_1, a_2, \dots, a_n\) represent a permutation. So, in this case we would choose \(a_1, a_2, \dots, a_k.\)

There are a total of \(n!\) permutations of the \(n\) elements. However, we choose the same \(k\) elements as long as \(a_1, a_2, \dots, a_k\) are the first \(k\) elements and \(a_{k-1}, a_{k-2}, \dots, a_n\) are the last \(n-k\) elements. To cancel the repetition of the ways we choose the same \(k\) elements, we must divide by all permutations of the first \(k\) elements and all permutations of the last \(n-k\) elements.

There are \(k!\) permutations of the first \(k\) elements and \((n-k)!\) permutations of the last \(n-k\) elements. After cancelling the repetition, the total number of different ways to choose \(k\) elements from \(n\) is \[\frac{n!}{k!(n-k)!} = {n \choose k}\]

This proof relies on the multiplication rule for counting. Each choice we make will be an experiment.

1. Choose a first element. There are \(n\) choices.
2. Choose a second element. There are \(n-1\) choices remaining.
3. Choose a third element. There are \(n-2\) choices remaining.
\(\vdots\)
4. Choose a \(k\)th element. There are \(n-k+1\) choices remaining.

By the multiplication rule, there are \(n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot (n-k+1)\) ways to choose the first \(k\) elements in a particular order.

We do not want to count the elements in a particular order. So, we need to eliminate the repetition of choosing the same set of \(k\) elements in different orders. Each different order is a different permutation, and there are \(k!\) permutations of \(k\) elements. So, after eliminating repetition, the number of ways to choose \(k\) elements from \(n\) is \[\frac{n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot (n-k+1)}{k!}\] Next, multiply the numerator and denominator by \((n-k)!\) to write the number of choices as \({n \choose k}.\) \begin{align} \frac{n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot (n-k+1)}{k!} & = \frac{n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot (n-k+1) \cdot (n-k)!}{k!(n-k)!} \\ & = \frac{n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot 2 \cdot 1}{k!(n-k)!} \\ & = \frac{n!}{k!(n-k)!} \\ & = {n \choose k} \end{align}

Since the company is choosing \(3\) colors from a set of \(10,\) the number of possible color combinations is \(10\) choose \(3,\) or \({10 \choose 3}.\) \[{10 \choose 3} = \frac{10!}{3!7!} = \frac{8 \cdot 9 \cdot 10}{1 \cdot 2 \cdot 3} = 120\] There are 120 possible color combinations.