Solving a Combinatorial Problem: Arrangements with Restrictions

  • Thread starter Thread starter Nathew
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary
When arranging the letters A, B, C, D, E, and F with the condition that A must come before B, the total number of valid arrangements is indeed half of the total arrangements without restrictions. This is because for every arrangement where A is before B, there is a corresponding arrangement where B is before A. The confusion arises when calculating specific cases, as demonstrated by the attempt to break down the problem using smaller sets of letters. The calculations provided, such as 2(5!)+2(4!)(2!)+(3!)(3!), suggest a misunderstanding of the combinatorial principles at play. Ultimately, the correct approach confirms that the arrangements with A before B should equal half of the unrestricted arrangements.
Nathew
If we have the letters A, B, C, D, E, and F, and we are asked to find the number of arrangements where A is before B, wouldn't this just be half of the total number of arrangements with no restrictions? Intuitively this makes sense, but I have some doubts. For example, when I try to do the problem out, I get 2(5!)+2(4!)(2!)+(3!)(3!) which is slightly over half of the total arrangements with no restrictions.

Help me out please!
 
Physics news on Phys.org
Try it for a smaller number of letters so that you can write out all the possibilities.
 

Thread 'Hypothesis testing: Defining H0, HA hypotheses so that ( H_A)_A' makes sense'

The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...
View full post »

Similar threads

High School Probability of arranging identical objects
  • · Replies 3 ·
Replies
3
Views
2K
High School Really simple mistake in this combinatorics problem?
  • · Replies 19 ·
Replies
19
Views
3K
Undergrad What is the probability that the product of the digits is odd?
  • · Replies 13 ·
Replies
13
Views
2K
Undergrad How Are Permutations Calculated in a Circular Seating Arrangement?
  • · Replies 2 ·
Replies
2
Views
2K
Counting Quadruples of Integers: A Combinatorial Problem
  • · Replies 3 ·
Replies
3
Views
2K
MHB How many words can be formed from the word SUCCESS without any repeated letters?
  • · Replies 2 ·
Replies
2
Views
2K
High School Baccarat math and beating the edge
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Ways to put n balls in m boxes such that exactly k boxes are empty?
  • · Replies 29 ·
Replies
29
Views
4K
MHB Number of ways to arrange letters
  • · Replies 3 ·
Replies
3
Views
2K
Number of ways of arranging 7 characters in 7 spaces
  • · Replies 8 ·
Replies
8
Views
2K