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Ksivusya [100]
3 years ago
6

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 i

tems, and list each of the permutations of items B, D, and F.

Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0

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.

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================================================

Explanation:

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