 
*P*(7, 4) · *P*(6, 3)

*P*(5, 4)

The expression you've provided involves permutations. The permutation formula is defined as:

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

where:
- \( n \) is the total number of items.
- \( r \) is the number of items to be chosen.

Let's break down each part of the expression:

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

\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\]

\[
3! = 3 \times 2 \times 1 = 6
\]

\[
P(7, 4) = \frac{5040}{6} = 840
\]

### 2. \( P(6, 3) \)
\[
P(6, 3) = \frac{6!}{(6 - 3)!} = \frac{6!}{3!}
\]

\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

\[
P(6, 3) = \frac{720}{6} = 120
\]

### 3. \( P(5, 4) \)
\[
P(5, 4) = \frac{5!}{(5 - 4)!} = \frac{5!}{1!}
\]

\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

\[
1! = 1
\]

\[
P(5, 4) = \frac{120}{1} = 120
\]

### Now, combine them as per the given expression:

\[
P(7, 4) \times P(6, 3) = 840 \times 120 = 100800
\]

\[
\frac{100800}{P(5, 4)} = \frac{100800}{120} = 840
\]

So, the final value of the expression \( P(7, 4) \cdot P(6, 3) / P(5, 4) \) is **840**.

Would you like further details or explanations? Here are five related questions for further practice:

1. What is \( P(8, 5) \) and how does it compare to \( P(7, 4) \)?
2. How would you simplify \( P(7, 4) \cdot P(6, 3) \cdot P(5, 2) \)?
3. How does \( P(n, r) \) change if you increase \( n \) by 1 but keep \( r \) the same?
4. How can you express a combination \( C(n, r) \) in terms of permutations?
5. How do permutations differ from combinations in terms of order and selection?

**Tip:** Remember that permutations consider the order of selection, making them useful in scenarios where the sequence matters.

Learn how to calculate permutations P(7, 4), P(6, 3), and P(5, 4) with detailed step-by-step solutions. Understand the permutation formula and its application. Suitable for advanced high school students and those preparing for competitive exams.

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