Let's solve both parts step by step.

Part (a): Ranking the Top 5 Salespeople from 31

Problem:
A board member wants to rank the top 5 salespeople in order of effectiveness from a total of 31 salespeople. How many rankings are possible?

Solution:
This is a permutation problem because the order of ranking matters. To determine how many ways we can rank 5 salespeople out of 31, we use the permutation formula:

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

where:

• $n = 31$ (total number of salespeople),
• $r = 5$ (we are choosing 5 people to rank in order).

$P(31, 5) = \frac{31!}{(31 - 5)!} = \frac{31!}{26!}$

This simplifies to:

$P(31, 5) = 31 \times 30 \times 29 \times 28 \times 27$

Now calculate:

$31 \times 30 = 930$ $930 \times 29 = 26,970$ $26,970 \times 28 = 755,160$ $755,160 \times 27 = 20,389,320$

Answer for (a): There are 20,389,320 possible rankings.

Part (b): Choosing 3 Customers from 50

Problem:
From 50 store customers, 3 are to be chosen to receive a gift. How many groups of 3 customers are possible?

Solution:
This is a combination problem because the order does not matter. To determine how many ways we can choose 3 customers out of 50, we use the combination formula:

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

where:

• $n = 50$ (total customers),
• $r = 3$ (choosing 3 customers).

$C(50, 3) = \frac{50!}{3!(50 - 3)!} = \frac{50 \times 49 \times 48}{3 \times 2 \times 1}$

Simplify step by step:

1. Multiply the top numbers:

$50 \times 49 = 2,450$ $2,450 \times 48 = 117,600$

2. Divide by $3! = 6$:

$\frac{117,600}{6} = 19,600$

Answer for (b): There are 19,600 possible groups of 3 customers.

Final Answers:

1. (a): 20,389,320 possible rankings.
2. (b): 19,600 possible groups.

Let me know if you'd like further clarification or a breakdown of any step. Here are some related questions to explore:

1. How would the result change if order didn't matter in part (a)?
2. How can we extend the formula to handle larger group sizes?
3. What if, in part (b), we also ranked the chosen 3 customers?
4. How are permutations and combinations used in probability?
5. How do factorials grow as $n$ increases?

Tip: Always check whether order matters in counting problems—this determines whether to use permutations or combinations!