Let's evaluate the given expression step by step:

The expression is:

$\frac{12^{C}10}{12^{P}4}$

Step 1: Understand the notation

• $12^{C}10$ refers to the number of combinations of 12 objects taken 10 at a time. The formula for combinations is: $n^{C}r = \frac{n!}{r!(n - r)!}$ So, $12^{C}10 = \frac{12!}{10!(12 - 10)!} = \frac{12!}{10!2!}$

• $12^{P}4$ refers to the number of permutations of 12 objects taken 4 at a time. The formula for permutations is: $n^{P}r = \frac{n!}{(n - r)!}$ So, ( 12^{P}4 = \frac{12!}{(12 - 4)!} = \frac{12!}{8!} ]

Step 2: Compute the values

We will now compute each part of the expression.

For $12^{C}10$:

$12^{C}10 = \frac{12!}{10!2!} = \frac{12 \times 11 \times 10!}{10! \times 2!} = \frac{12 \times 11}{2!} = \frac{12 \times 11}{2} = \frac{132}{2} = 66$

For $12^{P}4$:

$12^{P}4 = \frac{12!}{8!} = 12 \times 11 \times 10 \times 9 = 11880$

Step 3: Compute the final expression

Now we compute:

$\frac{12^{C}10}{12^{P}4} = \frac{66}{11880}$

Simplifying the fraction:

$\frac{66}{11880} = 0.0056$

Final Answer:

The value of the expression is $0.0056$.

Would you like further details on any part of this calculation? Here are some related questions to expand on this topic:

1. What is the difference between combinations and permutations?
2. How would the result change if you swapped the values for combinations and permutations?
3. How do you calculate factorials efficiently for large numbers?
4. What are the practical applications of combinations and permutations?
5. How would you solve this problem if the numbers involved were larger?

Tip: Remember that combinations don't account for order, while permutations do. This is why the formulas differ!