# Permutations and Combinations Practice Problems

• Draggu
In summary, the conversation discusses the number of ways to choose a certain number of six letter subsets and student council members, taking into account permutations and combinations. The solutions are provided for each scenario.
Draggu
1. The problem statement, all variables and given/known

## Homework Statement

1)How many six letter subsets can you make if 2 are consonants and 4 are vowels.

2) There are 22 students in the student council and 4 people are to be elected. How many ways can they be elected if:
a) there must be a vice president and president
b) vice president is excluded

## The Attempt at a Solution

1)
21x21x5x5x5x5

Is there a way to do this via combinations?

2a) C(22,2) * C(2,2)
b) C(22,3) * C(2,1)

These are probably horribly wrong, but I am trying. Any help would be appreciated!

1) You have not allowed for permutations of the letters chosen. IF all 6 letters are different, then there are 6! ways but you will have to allow for the same two consonants, etc.

2) First, choose a president. There are 22 ways to do that. Then choose a vice-president. There are 21 ways to do that. Finally choose 2 other members from the 20 remaining students. Permutations of the same students are not relevant.

3) Again, there are 22 ways to choose a president. Then choose 3 other members from the 21 remaining students. Permutations of the same students are not relevant.

HallsofIvy said:
1) You have not allowed for permutations of the letters chosen. IF all 6 letters are different, then there are 6! ways but you will have to allow for the same two consonants, etc.

2) First, choose a president. There are 22 ways to do that. Then choose a vice-president. There are 21 ways to do that. Finally choose 2 other members from the 20 remaining students. Permutations of the same students are not relevant.

3) Again, there are 22 ways to choose a president. Then choose 3 other members from the 21 remaining students. Permutations of the same students are not relevant.

1) Not quite sure what you mean here.

2) (22,1) * (21,1) * (20,2) ?

3) (22,1) * (21,3)

Anyone? I have a test tomorrow and am unsure :/

## What are permutations?

Permutations are a way of arranging a set of objects in a particular order. It is important to note that the order matters when dealing with permutations.

## What is the formula for calculating permutations?

The formula for calculating permutations is nPr = n! / (n-r)!, where n is the total number of objects and r is the number of objects being selected for the permutation.

## How do I know when to use permutations?

Permutations are used when you need to determine the number of ways to arrange a set of objects in a specific order. This can be useful in various mathematical problems, such as probability and statistics.

## Can you provide an example of a permutation problem?

Sure! Let's say you have 6 different colored marbles and you want to arrange them in a line. The number of permutations would be 6P6 = 6! / (6-6)! = 6! / 0! = 6! = 720. This means there are 720 different ways to arrange the marbles in a line.

## What are some common mistakes to avoid when dealing with permutations?

One common mistake is confusing permutations with combinations. Combinations are used when the order does not matter, while permutations are used when the order does matter. It is also important to remember to use the correct formula and to correctly identify the values for n and r in the problem.

• Precalculus Mathematics Homework Help
Replies
5
Views
1K
• Precalculus Mathematics Homework Help
Replies
6
Views
2K
• Precalculus Mathematics Homework Help
Replies
20
Views
3K
• Precalculus Mathematics Homework Help
Replies
4
Views
2K
• General Math
Replies
1
Views
736
• Precalculus Mathematics Homework Help
Replies
10
Views
3K
• Precalculus Mathematics Homework Help
Replies
2
Views
4K
• Precalculus Mathematics Homework Help
Replies
12
Views
2K
• Precalculus Mathematics Homework Help
Replies
2
Views
1K
• Precalculus Mathematics Homework Help
Replies
5
Views
3K