Question

1. a science quiz has eight multiple choice questions with four choices each. find the total number of ways to answer the questions.
2. there are 16 skaters competing in the competition. how many ways can they win the gold, silver, and bronze medal?
3. there are 23 students in ms. malboeufs class. how many different 3 person groups can she make?

Explanation:

Question 4

Step1: Identify the number of choices per question

Each of the 8 multiple - choice questions has 4 choices.

Step2: Use the multiplication principle

For each question, there are 4 ways to answer it. Since there are 8 independent questions, the total number of ways to answer all the questions is $4\times4\times\cdots\times4$ (8 times). Using the formula for exponents, this is $4^{8}$.
Calculate $4^{8}=4\times4\times4\times4\times4\times4\times4\times4 = 65536$.

Step1: Understand the problem as a permutation

We need to find the number of ways to select 3 skaters out of 16 for gold, silver, and bronze medals. The order matters here (gold is different from silver and bronze), so we use the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 16$ and $r=3$.

Step2: Calculate the permutation

First, calculate $n!=16!$, $(n - r)!=(16 - 3)!=13!$. Then $P(16,3)=\frac{16!}{13!}=\frac{16\times15\times14\times13!}{13!}=16\times15\times14$.
Calculate $16\times15 = 240$, then $240\times14 = 3360$.

Step1: Understand the problem as a combination

We need to find the number of ways to select 3 students out of 23, and the order of selection does not matter. The combination formula is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 23$ and $r = 3$.

Step2: Calculate the combination

First, calculate $n!=23!$, $r!=3!$, $(n - r)!=(23 - 3)!=20!$. Then $C(23,3)=\frac{23!}{3!\times20!}=\frac{23\times22\times21\times20!}{3\times2\times1\times20!}$.
Simplify the expression: $\frac{23\times22\times21}{6}$. Calculate $23\times22 = 506$, $506\times21 = 10626$, then $\frac{10626}{6}=1771$.

Answer:

65536

Question 5