Respuesta :
Answer:
[tex]\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}[/tex]
Step-by-step explanation:
If the system contains [tex]n[/tex] atoms, we can arrange [tex]r[/tex] quanta of energy in
[tex]\binom{n}{r} = \dfrac{n!}{r!(n-r)!}[/tex]
ways.
[tex]\mathbf{a)}[/tex]
In this case,
[tex]n = 2, r=1[/tex].
Therefore,
[tex]\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2[/tex]
which means that we can arrange 1 quanta of energy in 2 ways.
[tex]\mathbf{b)}[/tex]
In this case,
[tex]n = 20, r=10[/tex].
Therefore,
[tex]\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756[/tex]
which means that we can arrange 10 quanta of energy in 184 756 ways.
[tex]\mathbf{c)}[/tex]
In this case,
[tex]n = 2 \times 10^{23}, r = 10^{23}[/tex].
Therefore, we obtain that the number of ways is
[tex]\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}[/tex]
