Question

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 11 different dogs, but you have room for only 4 of them. how many different dog families could you have?

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. permutations should be used because we make selections from a group of choices.
c. arrangements with repetition should be used because we make selections from a group of choices.
d. permutations should be used because no item may be selected more than once and the order matters.
e. combinations should be used because no item may be selected more than once and the order does not matter.

calculate how many different dog families are possible.

(type a whole number.)

Explanation:

Identify the counting scenario

We need to choose 4 dogs out of 11 different dogs. Since each dog is unique, no dog can be selected more than once. The order in which we select or house the dogs does not change the group of dogs we have (the "dog family"). Therefore, order does not matter.

Using the Combinations knowledge point, this scenario matches combinations because we are selecting a subset where order is irrelevant and repetition is not allowed.

Select the correct option

Using the Permutations and Arrangements with Repetitions knowledge points, we compare the choices:

• Option E states: "Combinations should be used because no item may be selected more than once and the order does not matter."

This matches our analysis perfectly.

Calculate the number of combinations

Using the Combinations knowledge point, we calculate the number of ways to choose 4 dogs from 11:

$$ C(11, 4) = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} $$

Simplify the calculation

We simplify the fraction:

$$ C(11, 4) = \frac{11 \times 10 \times 9 \times 8}{24} = 11 \times 10 \times 3 = 330 $$

Answer:

Question 1

• (A) Arrangements with repetition should be used because no item may be selected more than once and the order matters.
• (B) Permutations should be used because we make selections from a group of choices.
• (C) Arrangements with repetition should be used because we make selections from a group of choices.
• (D) Permutations should be used because no item may be selected more than once and the order matters.
• (E) Combinations should be used because no item may be selected more than once and the order does not matter. (Correct answer)

Question 2

The number of different dog families is <blank>330</blank>.