Probability of 2 items not next to each other

  • Thread starter Thread starter desmond iking
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary
SUMMARY

The discussion centers on the probability of arranging items such that two specific items are not adjacent. The correct calculation for part e of the homework is 504 - (7C1 x 4) = 476, as confirmed by multiple participants. The confusion arises from the misinterpretation of the arrangement possibilities, particularly regarding the placement of one chosen item from seven. The total arrangements of three items out of nine, including the specific texts, must account for adjacency, leading to the conclusion that the book's answer is accurate.

PREREQUISITES
  • Combinatorial mathematics, specifically combinations (e.g., 7C1)
  • Understanding of permutations and arrangements (e.g., 3!)
  • Basic probability concepts related to item placement
  • Familiarity with problem-solving in pure mathematics and statistics
NEXT STEPS
  • Study combinatorial counting techniques in depth
  • Learn about permutations and their applications in probability
  • Explore adjacency problems in combinatorial mathematics
  • Review examples of arranging items with restrictions in probability theory
USEFUL FOR

Students in mathematics, particularly those studying combinatorics and probability, as well as educators looking to clarify concepts related to item arrangement and adjacency in problem-solving scenarios.

desmond iking
Messages
284
Reaction score
2

Homework Statement



i am having problem on part e , the ans given is 504-(7c1 X4 )= 476

but my working would be 504-(7c1 X4 x2 )= 448

since the pure maths and statistics can be arranged in 2! ways , but the 1 tem which is chosen from 7 items can be placed in 4 location , just as in the picture.



Homework Equations





The Attempt at a Solution

 

Attachments

  • kk.png
    kk.png
    3.2 KB · Views: 497
  • IMG_20140724_014823[1].jpg
    IMG_20140724_014823[1].jpg
    46 KB · Views: 464
Physics news on Phys.org
I have no idea what you mean by "but the 1 tem which is chosen from 7 items can be placed in 4 locations". (c) and so (e) refer to choosing 3 items out of 9. There are NO "4 locations".
How many different ways are there to choose 3 items out of 9 and then place them in some order? (i.e. part (c)). How many of those involve choosing the pure math and statistics text. How many of those have the two texts as first and second or as second and third? Subtract the second answer from the first.
 
  • Like
Likes   Reactions: 1 person
The answer in the book is correct.

Your "4x2" doesn't make sense. Consider three items, labeled A, B, and C, and arranged in a line. Some of the arrangements have items A and B adjacent, others don't. The total number of arrangements is 3!=6. The number in which items A and B are adjacent has to be less than this.
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
18
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K