The problem in the image asks the following:

How many permutations of three items can be selected from a group of six?

You are also asked to:

• Use the letters A, B, C, D, E, and F to identify the items.
• List all permutations of the specific items B, D, and F.

Step 1: Find the number of permutations

The formula for finding the number of permutations of $r$ items from a set of $n$ items is given by:

$P(n, r) = \frac{n!}{(n - r)!}$

In this case:

• $n = 6$ (since you have 6 items: A, B, C, D, E, F)
• $r = 3$ (you are choosing 3 items)

So, the number of permutations is:

$P(6, 3) = \frac{6!}{(6 - 3)!} = \frac{6 \times 5 \times 4 \times 3!}{3!} = 6 \times 5 \times 4 = 120$

Thus, there are 120 permutations of three items from a group of six.

Step 2: List the permutations of B, D, and F

The number of permutations of the specific items B, D, F is:

$3! = 3 \times 2 \times 1 = 6$

These six permutations are:

1. BDF
2. BFD
3. DBF
4. DFB
5. FBD
6. FDB

Summary:

• There are 120 total permutations of any three items selected from the six.
• The six permutations of B, D, and F are: BDF, BFD, DBF, DFB, FBD, and FDB.

Do you have any questions or would you like more details?

Related Questions:

1. How many combinations are there when selecting 3 items from 6 without considering the order?
2. What is the difference between permutations and combinations?
3. How would the solution change if you were asked to permute 4 items from a group of 6?
4. Can you calculate the total number of permutations for 5 items from a set of 7?
5. What are the permutations if the selected letters were A, B, and C instead of B, D, and F?

Tip:

For permutation problems, always keep in mind that the order matters—this distinguishes them from combinations.