Find all combinations of 6 items. Any order, no repeats.

  • Context: Undergrad 
  • Thread starter Thread starter Xzavios
  • Start date Start date
  • Tags Tags
    Combinations
Click For Summary
SUMMARY

The discussion centers on calculating the number of combinations of 6 distinct items (ABCDEF) without repetition and where order does not matter. The solution employs binomial coefficients and the formula 2^n - 1, where n is the number of items. For 6 items, this results in 63 combinations, accounting for all non-empty subsets. The user also highlights the practical application of this calculation in generating images for a client.

PREREQUISITES
  • Understanding of binomial coefficients
  • Familiarity with combinatorial mathematics
  • Basic knowledge of set theory
  • Experience with practical applications of combinatorial calculations
NEXT STEPS
  • Study binomial coefficients in depth
  • Learn about combinatorial mathematics and its applications
  • Explore set theory concepts related to combinations and permutations
  • Investigate practical uses of combinatorial calculations in design and data visualization
USEFUL FOR

This discussion is beneficial for mathematicians, data analysts, designers involved in data visualization, and anyone interested in combinatorial mathematics for practical applications.

Xzavios
Messages
2
Reaction score
0
This may be simple but I want to see a formula and admit I've been out of school too long to figure this one out. Although it's bugging me! I would like to see a formula, too, not just the answer.

So you have 6 items. For example ABCDEF.
You must use at least one item.
Order does not matter (ie. ABC=ACB)
They can not repeat.

So how many different combinations can I create with these items?

Thanks in advanced!
 
Physics news on Phys.org
1 item - 6, 2 items - 6x5/2, 3 items - 6x5x4/(3x2), 4 items - same as 2, 5 items - same as 1, 6 items - 1.

It is basically binomial coefficients.
 
If you don't care to break out the number of combinations based on the number of items in each group (as done by mathman), it's even simpler. Each item either is or is not included, so there are 2^6 possibilities. But you want at least one item in each selection, so subtract 1 to account for the empty set. Solution: 2^6 - 1.
 
Thanks guys I knew it was something so simple. As a designer I don't use math much anymore although it was always one of my favorite subjects in school.

This is great because this was based on a real life situation of creating images of these groups for a client. I created 60 so I see now that I'm missing 3.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K