Question

How many permutations of three items can be selected from a group of six? Use the letters A, B, C, D, E, and F to identify the items, and list each of the permutations of items B, D, and F.

Answer 1

Answer:

In total, $120$ permutations of three items can be selected from a group of six distinct elements.

In particular, there are $6$ ways to order three distinct items.

$\begin{aligned}\rm B-D-F \\ \rm B-F-D \\ \rm D-B-F \\ \rm D-F-B \\ \rm F-B-D \\ \rm F-D-B\end{aligned}$.

Step-by-step explanation:

The formula $\displaystyle P(n,\, r) = \frac{n!}{(n - r)!} = n \, (n - 1) \cdots (n - r + 1)$ gives the number of ways to select and order $r$ items from a group of $n$ distinct elements.

To select and order three items from a group six distinct elements, let $n = 6$ and $r = 3$. Apply the formula:

$\begin{aligned} P(6,\, 3) &= \frac{6!}{(6 - 3)!} = \frac{6!}{3!} \\ &= \frac{6 \times 5 \times 4 \times 3\times 2 \times 1}{3 \times 2 \times 1} \\ &= 6 \times 5 \times 4 = 120 \end{aligned}$.

In other words, there are $120$ unique ways to select and order three items (select a permutation of three items) from a group of six distinct elements.

Consider: what's the number of ways to order three distinct items? That's the same as asking: how many ways are there to select and order three items from a group of three distinct elements? Let $n =3$ and $r = 3$. Apply the formula for permutation:

$\begin{aligned} P(3,\, 3) &= \frac{3!}{(3 - 3)!} = \frac{3!}{0!} && \left(\text{ 0! = 1 by convention.}\right) \\ &= 3! = 3 \times 2\times 1 \\ &= 6\end{aligned}$.

To find the permutations, start by selecting one element as the first of the list. A tree diagram might be helpful. Refer to the attachment for an example.