From a set of 6 distinct items, in how many different ways can ordered arrangements of 4 items be formed (that is, in how many ways can we select groups of 4 at a time where order matters)?

Introduction / Context:
This problem is another example of permutations, where we must form ordered arrangements from a larger set. We have 6 distinct items and wish to create sequences of length 4 using these items without repetition. Because order matters, each different ordering of the same 4 items counts as a different arrangement.

Given Data / Assumptions:

• Total items available = 6.
• We form ordered groups of size 4.
• No item is repeated within a single arrangement.
• Different orders of the same 4 items are counted as different arrangements.

Concept / Approach:
The number of ordered arrangements of r items chosen from n distinct items is given by the permutation formula P(n, r) = n! / (n - r)!. Here, n = 6 and r = 4. We can plug these directly into the formula or think in terms of choices for each position in the ordered arrangement and multiply the number of choices.

Step-by-Step Solution:
Step 1: Apply the permutation formula. P(6, 4) = 6! / (6 - 4)! = 6! / 2!. 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. 2! = 2. So P(6, 4) = 720 / 2 = 360. Step 2: Alternatively, count by positions. First position: 6 choices. Second position: 5 choices (after one item is used). Third position: 4 choices. Fourth position: 3 choices. Total arrangements = 6 * 5 * 4 * 3 = 360.

Verification / Alternative check:
The two methods 6! / 2! and 6 * 5 * 4 * 3 agree, both giving 360. Listing all permutations is possible but unnecessary and time consuming. The standard permutation formula is sufficient to verify that 360 is the correct count for the number of ordered arrangements of 4 items from 6 distinct items.

Why Other Options Are Wrong:
The value 720 equals 6! and would correspond to arranging all 6 items, not just 4 at a time. The value 640 does not result from any natural factorial based calculation here. The value 740 is close to 720 but is not a factorial or permutation count. The number 120 equals 5! and is too small for P(6, 4). None of these values matches 6 * 5 * 4 * 3.

Common Pitfalls:
A common mistake is to treat the problem as combinations, counting only C(6, 4) groups and ignoring order. Another error is to miscalculate 6! or forget to divide by 2!. Some might also incorrectly think that all 6 items must be used and use 6! directly. Being clear that we want ordered groups of length 4 and applying P(6, 4) = 6! / 2! ensures the correct approach.

Final Answer:
There are 360 different ordered arrangements of 4 items from the set of 6.