Let's go step by step to solve each of the three parts.

a. $\frac{21!}{18!}$

Using the property of factorials:

$\frac{21!}{18!} = \frac{21 \times 20 \times 19 \times 18!}{18!}$

The $18!$ terms cancel out, leaving:

$21 \times 20 \times 19 = 7980$

Thus, $\frac{21!}{18!} = 7980$.

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b. $7P_5$ (Permutation of 7 objects taken 5 at a time)

The formula for permutation is:

$nP_r = \frac{n!}{(n-r)!}$

For $7P_5$:

$7P_5 = \frac{7!}{(7-5)!} = \frac{7!}{2!} = \frac{5040}{2} = 2520$

Thus, $7P_5 = 2520$.

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c. $\binom{10}{7}$ (Combination of 10 objects taken 7 at a time)

The formula for combination is:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

For $\binom{10}{7}$:

$\binom{10}{7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$

Thus, $\binom{10}{7} = 120$.

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Final Answers:

• a. $\frac{21!}{18!} = 7980$
• b. $7P_5 = 2520$
• c. $\binom{10}{7} = 120$

Would you like further details or have any questions?

Here are five related questions to explore further:

1. What is the difference between permutations and combinations?
2. How does the factorial function work in more general cases?
3. Can you calculate $10P_4$?
4. What would $\binom{12}{5}$ be?
5. How do we interpret these results in real-world problems?

Tip: Remember, permutations are used when order matters, while combinations are used when order does not matter.