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:In total, [tex]120[/tex] permutations of three items can be selected from a group of six distinct elements.In particular, there are [tex]6[/tex] ways to order three distinct items.[tex]\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}[/tex].Step-by-step explanation:The formula [tex]\displaystyle P(n,\, r) = \frac{n!}{(n - r)!} = n \, (n - 1) \cdots (n - r + 1)[/tex] gives the number of ways to select and order [tex]r[/tex] items from a group of [tex]n[/tex] distinct elements.To select and order three items from a group six distinct elements, let [tex]n = 6[/tex] and [tex]r = 3[/tex]. Apply the formula:[tex]\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}[/tex].In other words, there are [tex]120[/tex] 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 [tex]n =3[/tex] and [tex]r = 3[/tex]. Apply the formula for permutation:[tex]\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}[/tex].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.