Solve the following: a.
21!/18
! b.
7P 
5
​
 
c.
10
​
 C 
7
​


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, \( \frac{21!}{18!} = 7980 \).

---

### 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, \( 7P_5 = 2520 \).

---

### 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, \( \binom{10}{7} = 120 \).

---

### 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.

Solve the following: a. 21!/18! b. 7P_5 c. 10C_7

Discover the step-by-step solution to solving factorial, permutation, and combination problems: 21!/18!, 7P_5, and 10C_7. Understand the key concepts of factorials, permutations, and combinations with detailed explanations. Suitable for Grades 9-12.

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