Solution 1.3.5.1.

  1. You must simply choose 6 friends from a group of 14. This can be done in \({14 \choose 6}\) ways. We can find this number either by using Pascal's triangle or the closed formula: \(\frac{14!}{8!\cdot 6!} = 3003\text{.}\)

  2. Here you must count all the ways you can permute 6 friends chosen from a group of 14. So the answer is \(P(14, 6)\text{,}\) which can be calculated as \(\frac{14!}{8!} = 2162160\text{.}\)

    Notice that we can think of this counting problem as a question about counting functions: how many injective functions are there from your set of 6 chairs to your set of 14 friends (the functions are injective because you can't have a single chair go to two of your friends).

How are these numbers related? Notice that \(P(14,6)\) is much larger than \({14 \choose 6}\text{.}\) This makes sense. \({14 \choose 6}\) picks 6 friends, but \(P(14,6)\) arranges the 6 friends as well as picks them. In fact, we can say exactly how much larger \(P(14,6)\) is. In both counting problems we choose 6 out of 14 friends. For the first one, we stop there, at 3003 ways. But for the second counting problem, each of those 3003 choices of 6 friends can be arranged in exactly \(6!\) ways. So now we have \(3003\cdot 6!\) choices and that is exactly \(2162160\text{.}\)

Alternatively, look at the first problem another way. We want to select 6 out of 14 friends, but we do not care about the order they are selected in. To select 6 out of 14 friends, we might try this:

\begin{equation*} 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9\text{.} \end{equation*}

This is a reasonable guess, since we have 14 choices for the first guest, then 13 for the second, and so on. But the guess is wrong (in fact, that product is exactly \(2162160 = P(14,6)\)). It distinguishes between the different orders in which we could invite the guests. To correct for this, we could divide by the number of different arrangements of the 6 guests (so that all of these would count as just one outcome). There are precisely \(6!\) ways to arrange 6 guests, so the correct answer to the first question is

\begin{equation*} \frac{14 \cdot 13 \cdot 12 \cdot 11\cdot 10 \cdot 9}{6!}\text{.} \end{equation*}

Note that another way to write this is

\begin{equation*} \frac{14!}{8!\cdot 6!}\text{.} \end{equation*}

which is what we had originally.

in-context