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

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}`.

Explanation:

The formula
`\displaystyle P(n,\, r) = (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) &= (6!)/((6 - 3)!) = (6!)/(3!) \\ &= (6 * 5 * 4 * 3* 2 * 1)/(3 * 2 * 1) \\ &= 6 * 5 * 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) &= (3!)/((3 - 3)!) = (3!)/(0!) && \left(\text{$0! = 1$ by convention.}\right) \\ &= 3! = 3 * 2* 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.