Question

A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of energy when (a) n = 2, r = 1; (b) n = 20, r 10; (c) n = 2 x 1023, r = 1023?

Answer 1

Answer:

$\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

Step-by-step explanation:

If the system contains $n$ atoms, we can arrange $r$ quanta of energy in

                         $\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

ways.

$\mathbf{a)}$

In this case,

                                $n = 2, r=1$.

Therefore,

                    $\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2$

which means that we can arrange 1 quanta of energy in 2 ways.

$\mathbf{b)}$

In this case,

                                $n = 20, r=10$.

Therefore,

                    $\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756$

which means that we can arrange 10 quanta of energy in 184 756 ways.

$\mathbf{c)}$

In this case,

                                $n = 2 \times 10^{23}, r = 10^{23}$.

Therefore, we obtain that the number of ways is

                    $\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$