Distributing 6 Distinct Molecules Across 3 Energy Levels

  • Thread starter Thread starter patcho
  • Start date Start date
  • Tags Tags
    Molecules
Click For Summary
SUMMARY

The discussion focuses on calculating the number of ways to distribute 6 distinguishable molecules across 3 energy levels, specifically with 3 molecules in the first level, 2 in the second, and 1 in the third. The correct approach involves using combinations, leading to the formula 6C3 * 3C2 * 1C1, which results in 60 distinct arrangements. The breakdown of the calculation confirms that the initial assumption of 10 was incorrect, emphasizing the importance of applying the combination formula accurately.

PREREQUISITES
  • Understanding of combinations, specifically the formula nCr = n! / (r! * (n-r)!)
  • Familiarity with factorial notation and its application in combinatorial problems
  • Basic knowledge of distinguishable objects in probability and statistics
  • Concept of grouping and partitioning in combinatorial mathematics
NEXT STEPS
  • Study advanced combinatorial techniques, including permutations and combinations
  • Explore applications of combinatorial mathematics in statistical mechanics
  • Learn about multinomial coefficients and their relevance in distributing objects
  • Investigate the principles of partition theory in combinatorial analysis
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on combinatorics, as well as researchers in fields requiring statistical distribution analysis.

patcho
Messages
5
Reaction score
0
Hi,

I'm not quite sure about this question:

'How many Ways can 6 distinguishable molecules be placed in 3 different energy levels with 3 molecules in the 1st level, 2 in the 2nd level and 1 in the 3rd level, ignoring energy required?'

If it was just how many ways to place them in 3 different levels it would be easy but how to always keep 3 molecules in the first, 2 in the 2nd and 1 in the first confuses me.

I know that the number of Ways is less than before and I'm thinking along the lines of having to divide the number obtained if it was just 3 different energy levels, by 3!2!1! (the number of molecules in each level). This gives: W=10 which I think is very wrong!

Any help appreciated!

:smile:
 
Physics news on Phys.org
Take a slightly more general question:

How many ways can N numbered balls be grouped into three sets such that the first group contains [itex]n_1[/itex] balls, the second set [itex]n_2[/itex] balls and the third set [itex]N-n_1-n_2[/itex] balls?

Hint: How many ways are there to choose the first set? After that, in how many ways can you choose the second set?
 


Hi there,

Thank you for your question. It seems like you are trying to figure out the number of possible arrangements for 6 distinguishable molecules in 3 different energy levels, with specific numbers of molecules in each level.

In this case, you can use the concept of combinations to find the number of ways to place the molecules. The formula for combinations is nCr = n! / (r! * (n-r)!), where n is the total number of objects and r is the number of objects being chosen.

In this problem, we have 6 molecules and we want to choose 3 for the first level, 2 for the second level, and 1 for the third level. So, the number of ways to place these molecules would be:

6C3 * 3C2 * 1C1 = (6! / (3! * (6-3)!) * (3! / (2! * (3-2)!) * (1! / (1! * (1-1)!)
= (6*5*4 / (3*2*1)) * (3*2 / (2*1)) * (1 / (1*0))
= 20 * 3 * 1
= 60

So, there are 60 ways to place 6 distinguishable molecules in 3 different energy levels with 3 molecules in the first level, 2 in the second level, and 1 in the third level.

I hope this helps clarify the concept for you. If you have any further questions, please feel free to ask. Best of luck with your studies!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 22 ·
Replies
22
Views
2K
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
6K
  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 39 ·
2
Replies
39
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K