Permutations and Combinations Practice Problems

Click For Summary

Homework Help Overview

The discussion revolves around problems related to permutations and combinations, specifically focusing on forming subsets of letters and electing members from a student council. The original poster presents two distinct problems involving the selection of consonants and vowels, as well as the election of council members with specific roles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of combinations and permutations in solving the problems. The original poster questions whether combinations can be applied to their first problem and expresses uncertainty about their attempts. Others provide insights into the need to consider permutations of letters and suggest methods for electing council members.

Discussion Status

The discussion is ongoing, with participants offering various interpretations and approaches to the problems. Some guidance has been provided regarding the need to account for permutations in the first problem and the steps for electing council members, but no consensus has been reached on the correct methods or solutions.

Contextual Notes

The original poster indicates a time constraint due to an upcoming test, which may influence the urgency and nature of the responses sought in the discussion.

Draggu
Messages
101
Reaction score
0
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


Homework Equations





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!
 
Physics news on Phys.org


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 :/
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K