Question

unit 1 probability review
the following problems require you to determine the total possible outcomes of each event. for each problem, identify if the prompt is a fundamental counting principle problem, or if it uses a permutation or a combination. then, solve the problem.

1. there are 20 players on an nhl team, 2 of which are goalies. how many ways are there to choose 5 skaters and 1 goalie to be out on the ice?

type of problem:
solution: 12176

1. you are going on a road trip with 4 friends in a car that seats 5 people. how many ways are there to choose where everyone sits if you are always the driver?

type of problem:
solution:

1. twenty students are participating in a tennis tournament. how many ways are there for them to take 1st 2nd place?

type of problem:
solution:

1. how many ways are there to choose a six - digit passcode for your phone?

type of problem:

Explanation:

Step1: Calculate number of skaters

There are 20 - 2=18 skaters. We need to choose 5 skaters out of 18. The number of ways to do this is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 18$ and $r=5$. So, $C(18,5)=\frac{18!}{5!(18 - 5)!}=\frac{18\times17\times16\times15\times14}{5\times4\times3\times2\times1}=8568$.

Step2: Calculate number of goalies

There are 2 goalies and we need to choose 1. Using the combination formula with $n = 2$ and $r = 1$, $C(2,1)=\frac{2!}{1!(2 - 1)!}=2$.

Step3: Use fundamental counting principle

The total number of ways to choose 5 skaters and 1 goalie is the product of the number of ways to choose skaters and goalies. So the total number of ways is $8568\times2 = 17136$.

2.

Step1: Identify the problem type

Since you are always the driver, we need to arrange 4 friends in the remaining 4 seats. This is a permutation problem as the order of seating matters. The number of permutations of $n$ distinct objects taken $n$ at a time is given by $P(n,n)=\frac{n!}{(n - n)!}=n!$. Here $n = 4$.

Step2: Calculate the number of arrangements

$P(4,4)=\frac{4!}{(4 - 4)!}=4!=4\times3\times2\times1 = 24$.

We want to find the number of ways 20 students can take 1st and 2nd place. The order matters (1st - 2nd is different from 2nd - 1st). The number of permutations of 20 students taken 2 at a time is $P(20,2)=\frac{20!}{(20 - 2)!}=\frac{20!}{18!}=20\times19 = 380$.
Solution: 380
4.
Type of Problem: Fundamental counting principle

For each digit of the six - digit passcode, there are 10 possible choices (0 - 9). By the fundamental counting principle, the total number of six - digit passcodes is $10\times10\times10\times10\times10\times10=10^6 = 1000000$.
Solution: 1000000

Answer:

1.
Type of Problem: Combination and fundamental counting principle
Solution: 17136
2.
Type of Problem: Permutation
Solution: 24
3.
Type of Problem: Permutation