Combination Formula with a lockout twist

Click For Summary
SUMMARY

The discussion centers on calculating combinations of 6 items selected from 18 choices, divided into 6 groups with 3 options each. The key constraint is that once an item is selected from a group, the other two items in that group are "locked out." The established formula for this scenario is 3^6, indicating that there are 729 unique combinations possible. This approach mirrors the mechanics of independent selections, similar to the popular bike lock combinations.

PREREQUISITES
  • Understanding of combinatorial mathematics
  • Familiarity with the concept of "lockout" in selection processes
  • Basic knowledge of exponential functions
  • Ability to visualize group-based selection scenarios
NEXT STEPS
  • Explore advanced combinatorial algorithms for larger datasets
  • Learn about the application of combinatorial mathematics in cryptography
  • Investigate variations of the lockout principle in game theory
  • Study the implications of independent selections in probability theory
USEFUL FOR

Mathematicians, computer scientists, game designers, and anyone interested in combinatorial optimization and selection theory.

Calcunaut
Messages
2
Reaction score
0
Combination Formula with a "lockout" twist

Hi! I am trying to figure out all possible combonations for 6 items among a group of 18 choices. So I turn to my old friend C(n,r) to calculate where n=18 and r=6. "But WAIT!" I tell you before you hastily begin scribbling, "There is a twist..." You see my problem is that the items are divided up into 6 groups, with 3 choices in each group. Once a choice has been made in a group for the combination the other 2 in the group are unavailable, or "locked out" of the rest of the combination. The order doesn't necessarily matter but a choice must be selected from each of the six groups. Here's a visual representation:

A B C
1 A1 B1 C1
2 A2 B2 C2
3 A3 B3 C3
4 A4 B4 C4
5 A5 B5 C5
6 A6 B6 C6

If "B1" is selected in a single combination then "A1" and "C1" cannot be apart of the same combination. What is the formula for this and how many possible combinations are there?
 
Physics news on Phys.org


Do you have 18 distinct items? The answer is 3^6.

Or are you counting {A1,A2,A3,B4,B5,A6} to be the same as {A1,B2,B3,A4,A5,A6}?
 


If its like the bike locks that are popular, each "ring" is independent and like Stephen said, there will be 3x3x3x3x3x3 combinations (or 3^6).
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
8K
  • · Replies 2 ·
Replies
2
Views
4K