5. a contractor is installing a new door. there are six styles of doors, three different finishes, and eight door handles to choose from. how many doors are possible?
type __________
explanation: __________

6. samson is the hiring manager at costco. she needs to hire 5 new employees. twenty seven people have applied to work at costco. in how many ways can she fill the five new positions?
type __________
explanation: __________

7. you roll a six - sided die three times. how many outcomes are there?
type __________
explanation: __________

8. selina has x items in her lunch box. she is going to pull out one item from her lunch box. is this a permutation or a combination? explain.

Question

5. a contractor is installing a new door. there are six styles of doors, three different finishes, and eight door handles to choose from. how many doors are possible?
type __________
explanation: __________

6. samson is the hiring manager at costco. she needs to hire 5 new employees. twenty seven people have applied to work at costco. in how many ways can she fill the five new positions?
type __________
explanation: __________

7. you roll a six - sided die three times. how many outcomes are there?
type __________
explanation: __________

8. selina has x items in her lunch box. she is going to pull out one item from her lunch box. is this a permutation or a combination? explain.

Accepted Answer

### Problem 5 (Door Installation)
# Explanation:
## Step1: Identify the principle
We use the multiplication principle of counting. If there are $ n_1 $ ways to do one thing, $ n_2 $ ways to do a second thing, and $ n_3 $ ways to do a third thing, the total number of ways to do all three things together is $ n_1\times n_2\times n_3 $. Here, $ n_1 = 6 $ (styles), $ n_2=3 $ (finishes), $ n_3 = 8 $ (handles).
## Step2: Calculate the total number
Multiply the number of options for each category: $ 6\times3\times8 $
First, $ 6\times3 = 18 $, then $ 18\times8=144 $
# Answer:
144

### Problem 6 (Hiring Employees)
# Explanation:
## Step1: Determine the type of problem
This is a permutation or combination problem. Since the order in which we hire employees (i.e., assigning different positions) might matter (assuming different positions are distinct), we use permutations. The formula for permutations is $ P(n,r)=\frac{n!}{(n - r)!} $, where $ n = 27 $ (total applicants) and $ r=5 $ (positions to fill).
## Step2: Calculate the permutation
$ P(27,5)=\frac{27!}{(27 - 5)!}=\frac{27!}{22!}=27\times26\times25\times24\times23 $
Calculate step - by - step:
$ 27\times26 = 702 $
$ 702\times25=17550 $
$ 17550\times24 = 421200 $
$ 421200\times23=9687600 $
# Answer:
9687600

### Problem 7 (Rolling a Die)
# Explanation:
## Step1: Identify the counting principle
When rolling a die three times, for each roll, there are 6 possible outcomes. By the multiplication principle, if we have $ n $ independent trials (rolls) each with $ k $ outcomes, the total number of outcomes is $ k^n $. Here, $ k = 6 $ (sides of the die) and $ n = 3 $ (number of rolls).
## Step2: Calculate the total number of outcomes
The total number of outcomes is $ 6^3=6\times6\times6 $
$ 6\times6 = 36 $, then $ 36\times6 = 216 $
# Answer:
216

### Problem 8 (Lunch Box Item)
# Brief Explanations:
When pulling out one item from the lunch box, the order of pulling out one item does not matter (since we are just selecting one item, there's no arrangement involved). A combination is a selection of items where order does not matter, and a permutation is a selection where order matters. Since we are only selecting one item, both permutation and combination would give the same result in terms of count, but conceptually, since order doesn't matter for a single - item selection, it can be considered a combination (or a permutation with $ r = 1 $, as $ P(n,1)=C(n,1)=n $).
# Answer:
This is a combination (or a permutation with $ r = 1 $). The reason is that when selecting one item, the order of selection does not matter. For a single - item selection, the number of permutations $ P(n,1)=\frac{n!}{(n - 1)!}=n $ and the number of combinations $ C(n,1)=\frac{n!}{1!(n - 1)!}=n $, so both give the same result, but the nature of the selection (no order importance for one item) aligns with the concept of a combination.

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QUESTION IMAGE

Question

Explanation:

Problem 5 (Door Installation)

Step1: Identify the principle

Step2: Calculate the total number

Step1: Determine the type of problem

Step2: Calculate the permutation

Step1: Identify the counting principle

Step2: Calculate the total number of outcomes

Answer:

Problem 6 (Hiring Employees)