The equation in the image is:

$P_5 - 3! + 0! + P_2 - P(2,3) + P(10,4)$

Let's break it down:

1. $P_5$: Likely refers to the permutation of 5 objects, $P(5,5)$, which equals $5!$.
2. $3!$: Factorial of 3, which equals $6$.
3. $0!$: Factorial of 0, which equals $1$.
4. $P_2$: Likely refers to $P(2,2) = 2!$.
5. $P(2,3)$: This seems incorrect because you cannot have 3 objects chosen from 2. I will assume it's an error.
6. $P(10,4)$: This is the number of ways to arrange 4 objects from 10, given by $\frac{10!}{(10-4)!} = \frac{10!}{6!}$.

Solving:

• $P_5 = 5! = 120$
• $3! = 6$
• $0! = 1$
• $P_2 = 2! = 2$
• $P(2,3)$ is invalid, but I'll treat it as 0 or possibly ignored.
• $P(10,4) = \frac{10!}{6!} = \frac{10 \times 9 \times 8 \times 7}{1} = 5040$

Now substitute into the equation:

$120 - 6 + 1 + 2 - 0 + 5040 = 5157$

So, the result is:

$5157$

Do you have any questions or would you like more details on a specific part?

Related Questions:

1. What is the difference between permutations and combinations?
2. How do you calculate $n!$ (factorial) for large values?
3. What is the general formula for permutations $P(n, r)$?
4. Why is $0! = 1$?
5. How can the value of $P(10, 4)$ be used in real-life applications?

Tip:

Factorials grow very quickly, so for large numbers, consider using logarithms or approximation techniques like Stirling's formula.