0
$\begingroup$

Hello MathStackExchange Community,

Confused about how to change the notation-formula for permutations when you have more complex processes. i.e. multiple independent action choices of different types of objects and orientation limitations.

  • The generic formula is : $P(n,r)= n!/(n-r)!$
  • $n!$ = being the total number of objects in an option set
  • $(n-r)!$ = the limitations caused by the choices made.

but this only works for very simple permutation-processes where you make one single choice-type.

If I have 15 particles, 10 neutral and 5 positive particles and I care about the arrangement

he formula is $P(n,r)= 15!/10!\cdot5!$
What I don't get is how does 10! 5! translate to $(n-r)!$

the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?

What is the way to think about the second portion? Could draw it out and try to find a pattern, but if I had avagadro's number of particles that would be hard.

Thanks,

ThermoRestart

$\endgroup$
0
$\begingroup$

What I don't get is how does 10! 5! translate to (n-r)!

You cannot see how the formula is applied because ... it is not being applied.   A slightly different formula is.

This is an application of the binomial coeffient, sometimes written $^n\mathrm C_r$, $\mathrm C(n,r)$, or as $\binom nr$.   It counts ways to select $r$ elements from a set of $n$ distinct elements.

$$^n\mathrm C_r = \dfrac{n!}{r!~(n-r)!}$$

In this case, we are counting ways to select places to put the $10$ neutral particles from the $15$ distinct places available. $$^{15}\mathrm C_{10}=\dfrac{15!}{10!~5!}$$


the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?

For this we have to know what shape the collections of fifteen particals has, such that "next to" has meaning.

Let us say we're stringing the particals in a chain. In that case, line the neutral particals up, and place each positive partical between two neutral particals, or at an end of the string (ie there are eleven posible places). $$\_\circ\_\circ\_\circ\_\circ\_\circ\_\circ\_\circ\_\circ\_\circ\_\circ\_$$ Count the ways.

$\endgroup$
  • $\begingroup$ Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible? $\endgroup$ – ThermoRestart Mar 20 '19 at 7:05
0
$\begingroup$

Limiting the orientation of particles limits the number of arrangements possible and makes the permutation counting concerned about the blocks of spaces that are cation or neutral instead of discrete spaces.

The correct number of permutations is 462.

If cation particles can't sit next to cation particles, that redefines the problem to consider spaces available instead of discrete particles. This is true because, the arrangements possible are completely limited by the number of cation particle spaces.

_ C _ C _ C_ C_C_ <--- describes the set up, where C= cation particle

this leaves 6 spaces for neutral particles.

The number of discrete neutral particles aren't considered because they don't limit the arrangement.

Apply the permutation formula for multiple choices.

N! = total objects available to choose from

n1! ... nt!= each a subset of indistinguishable objects, which were chosen from N

N_total_available_spaces!/n_neutral_spaces! n_cation_spaces!

11!/5!6!= 462

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.