No. of choosing 4 atoms out of 10 atoms

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

A) There is only one way to have 10 quanta of energy i.e. to have all atoms in higher states.
B) I have to decide no. of ways of assigning one units of energy to any 4 atoms. It's equivalent to choosing 4 atoms out of 10 atoms.
If the atoms are indistinguishable, the no. of ways of doing this is one .

If the atoms are distinguishable, then let's label each atom by one roman digits from i to x.
let's say that we have 4 different boxes A,B, C, D.
The total no. of ways in which we can put one atom in each box is 10*9*8*7.
When all of the boxes are indistinguishable,
let's consider one particular arrangement { (i, A) ,(ii, B ) ,( v,C) , (x,D)} among the 10*9*8*7 arrangements.
Since the boxes are indistinguishable, it doesn't matter whether the atom- i is in the box-A or box- B and so on.
When the boxes are distinguishable, there are 4*3*2*1 different ways of putting these four atoms into the four boxes.
When the boxes are made indistinguishable, these 4*3*2*1 different ways don't remain different. This happens with each set of four atoms in the 10*9*8*7 arrangements. Hence, the total no. of putting the atoms into four indistinguishable boxes is $\frac {10*9*8*7}{4*3*2*1}$.
Since this case corresponds to the problem asked, total no. of distinct arrangements = $\frac {10*9*8*7}{4*3*2*1}$.
Is this correct so far?

 

[andrewkirk]

Yes, that is correct.

 

[Pushoam]

Thank you.
In the question, it is not said that the atoms are indistinguishable. I am assuming it. Is the assumption correct?

 

[andrewkirk]

Since the question did not specify, they could be distinguishable or indistinguishable. It is insightful that you covered both cases. The distinguishable case would be where all atoms were different. In practice, it might be difficult to get ten distinguishable atoms in a box without them reacting with one another, as there are fewer than ten noble gases.

There are many other possibilities, for example three Neon atoms, three Argon and four Helium. That would give another, different calculation. It might be worth mentioning the other possibilities, but I wouldn't do the calcs for them. There are too many possibilities. You've covered the two most important ones.

 

[Pushoam]

There are many other possibilities, for example three Neon atoms, three Argon and four Helium.

Let's consider the case when four of the ten atoms are same,
no. of arrangements in which all of the four atoms are different is ${^7 C _4}$.
no. of arrangements in which three atoms are different is ${^6 C _2}$
no. of arrangements in which two atoms are different is ${^6 C _1}$
no. of arrangements in which none of the four atoms are different is ${^6 C _0}$.
So, total no. of arrangements is ${^7 C _4} + {^6 C _2}+{^6 C _1}+{^6 C _0}$.
Is this correct?

.

 

[andrewkirk]

Yes, that's correct.

 

[Pushoam]

Thank you for guiding me.

 

[CWatters]

A) There is only one way to have 10 quanta of energy i.e. to have all atoms in higher states.

That's correct if you have to use all the energy "at your disposal". What if you don't?

 

[Pushoam]

That's correct if you have to use all the energy "at your disposal". What if you don't?

That leads to a problem which is similar to the problem given in part (b), doesn't it?