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

`\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} ((2* 10^(23))!)/((10^(23)!)(10^(23))!)`

Explanation:

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

`\binom{n}{r} = (n!)/(r!(n-r)!)`

ways.

`\mathbf{a)}`

In this case,

`n = 2, r=1`.

Therefore,

`\binom{n}{r} = \binom{2}{1} = (2!)/(1!(2-1)!) = (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} = (20!)/(10!(20-10)!) = (10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20)/(10!10!) = (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 * 10^(23), r = 10^(23)`.

Therefore, we obtain that the number of ways is

`\binom{n}{r} = \binom{2* 10^(23)}{10^(23)} = ((2* 10^(23))!)/((10^(23))!(2* 10^(23) - 10^(23))!) = ((2* 10^(23))!)/((10^(23)!)(10^(23))!)`