A company is creating three new divisions and seven managers are eligible to be appointed head of a division.

How many different ways could the three new heads be appointed?

To determine the number of different ways the three new heads can be appointed from seven eligible managers, we need to consider the order in which the heads are appointed because each head position is distinct (e.g., Division A, Division B, Division C).

This is a permutation problem since the order matters. The formula for permutations is given by:

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

Where:
- \( n \) is the total number of items to choose from (7 managers),
- \( r \) is the number of items to choose (3 division heads).

Substituting the values:

\[
P(7, 3) = \frac{7!}{(7-3)!} = \frac{7!}{4!}
\]

Let's calculate this:

\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\]
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]
\[
P(7, 3) = \frac{5040}{24} = 210
\]

Thus, there are **210** different ways to appoint the three new heads.

Would you like any further details or explanations?

Here are some related questions to consider:
1. How would the solution change if order didn't matter?
2. What if one person could hold multiple head positions?
3. How many ways could the three heads be appointed if there were 10 eligible managers?
4. What if four heads needed to be appointed instead of three?
5. How does this problem relate to combinations versus permutations?

**Tip:** When the order matters in a selection process, always consider permutations instead of combinations.

Learn how to calculate permutations to appoint three division heads from seven eligible managers. Understand the permutation formula and step-by-step calculation. Suitable for advanced high school students.

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