Answer the following question using arrangements with​ repetition, permutations, or combinations. Be sure to explain why the particular counting technique applies to the problem.
A dog shelter is giving away 8 different​ dogs, but you have room for only 4 of them. How many different dog families could you​ have?
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Part 1
Determine the appropriate counting technique. Choose the correct answer below.
A.
Arrangements with repetition should be used because no item may be selected more than once and the order matters.
B.
Arrangements with repetition should be used because we make selections from a group of choices.
C.
Permutations should be used because no item may be selected more than once and the order matters.
D.
Combinations should be used because no item may be selected more than once and the order does not matter.
E.
Permutations should be used because we make selections from a group of choices.
Part 2
Calculate how many different dog families are possible.
  
enter your response here ​(Type a whole​ number.) 

### Part 1: Determining the Appropriate Counting Technique

The key points in the problem are:
- You are selecting 4 dogs out of 8 different dogs.
- Each dog can be selected only once (so no repetition).
- The order in which the dogs are selected does not matter (since the group of dogs is being referred to as a "family").

From these observations, the correct counting technique to apply is **combinations**, because:
- **No repetition**: Each dog can only be selected once.
- **Order does not matter**: It doesn't matter in which order the dogs are selected; the group of 4 dogs is what matters, not their arrangement.

Therefore, the correct answer is:
**D. Combinations should be used because no item may be selected more than once and the order does not matter.**

### Part 2: Calculating the Number of Different Dog Families

To calculate how many different dog families are possible, we use the combination formula:

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

Where:
- \( n = 8 \) (the total number of dogs),
- \( r = 4 \) (the number of dogs you are selecting).

Now, substitute these values into the combination formula:

\[
C(8, 4) = \frac{8!}{4!(8 - 4)!} = \frac{8!}{4!4!}
\]

Calculating factorials:

\[
8! = 8 \times 7 \times 6 \times 5 \times 4!
\]

\[
C(8, 4) = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = \frac{1680}{24} = 70
\]

Thus, the number of different dog families you can have is:

\[
\boxed{70}
\]

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Would you like any more details or clarifications?

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### Five Related Questions:
1. What is the difference between combinations and permutations?
2. How would the answer change if repetition of dogs was allowed?
3. What if the order of the dogs mattered? How would you calculate the number of arrangements then?
4. How many ways can you select 3 dogs from the 8?
5. What if you had to select 5 dogs instead of 4? How would that change the calculation?

### Tip:
When deciding between combinations and permutations, remember that combinations ignore order while permutations consider it.

A dog shelter is giving away 8 different​ dogs, but you have room for only 4 of them. How many different dog families could you​ have? Determine the appropriate counting technique and calculate how many different dog families are possible.

Learn how to calculate the number of different dog families you can form by selecting 4 dogs from 8 using combinations. This problem demonstrates the use of the combination formula, emphasizing that order does not matter. Suitable for students in Grades 9-12.

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