Let's break down the two expressions given:

(a) $^9P_2$:

This is a permutation problem, where we are asked to find the number of ways to arrange 2 objects out of 9. The formula for permutations is:

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

For $^9P_2$:

$^9P_2 = \frac{9!}{(9-2)!} = \frac{9!}{7!}$

This simplifies to:

$^9P_2 = 9 \times 8 = 72$

(b) $^9C_2$:

This is a combination problem, where we are asked to find the number of ways to choose 2 objects out of 9 without regard to the order. The formula for combinations is:

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

For $^9C_2$:

$^9C_2 = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!}$

This simplifies to:

$^9C_2 = \frac{9 \times 8}{2 \times 1} = \frac{72}{2} = 36$

Final Answers:

• (a) $^9P_2 = 72$
• (b) $^9C_2 = 36$

Let me know if you'd like any further clarification!

Here are 5 related questions to expand on these concepts:

1. What is the difference between permutations and combinations?
2. How would you compute $^8P_3$?
3. What is $^{10}C_4$ and how does it compare to $^{10}P_4$?
4. Can you explain how factorials are used in permutation and combination formulas?
5. In what types of problems would you use combinations instead of permutations?

Tip: When determining if a problem involves permutations or combinations, ask yourself if order matters. If it does, use permutations!