91Ó°ÊÓ

The Stirling’s Formula is:

If\({s_n} = \frac{1}{2}\log \left( {2\pi } \right) + \left( {n + \frac{1}{2}} \right)\log \left( n \right) - n\)then,\(\mathop {\lim }\limits_{n \to \infty } \left| {{s_n} - \log \left( {n!} \right)} \right| = 0\)

In other ways, \(\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\pi } \right)}^{\frac{1}{2}}}{n^{\frac{{n + 1}}{2}}}{e^{ - n}}}}{{n!}} = 1\)

If n and k both are positive integers,then,

\({P_{n,k}} = \frac{{n!}}{{k!\left( {n - k} \right)!}}\)

Assuming that n and (n-k) both are positive integers and large.

Using Stirling’s Formula, we get,

\(n! \approx {\left( {2\pi } \right)^{\frac{1}{2}}}{n^{n + \frac{1}{2}}}{e^{ - n}}\) and

\(\left( {n - k} \right)! \approx {\left( {2\pi } \right)^{\frac{1}{2}}}{\left( {n - k} \right)^{\left( {n - k} \right) + \frac{1}{2}}}{e^{ - n + k}}\)

Since the ratio of approximation of \(n!\)and \(\left( {n - k} \right)!\)converges to 1 to the corresponding factors of \(n!\)and \(\left( {n - k} \right)!\)

So,

\(\begin{aligned}{c}{P_{n,k}} = \frac{{n!}}{{k!\left( {n - k} \right)!}}\\ \approx \frac{{{{\left( {2\pi } \right)}^{\frac{1}{2}}}{n^{n + \frac{1}{2}}}{e^{ - n}}}}{{{{\left( {2\pi } \right)}^{\frac{1}{2}}}{{\left( {n - k} \right)}^{\left( {n - k} \right) + \frac{1}{2}}}{e^{ - n + k}}}}\\ = \frac{{{e^{ - k}}{n^k}}}{{k!}}{\left( {1 - \frac{k}{n}} \right)^{ - n - k - \frac{1}{2}}}\end{aligned}\)

Let, \(k < n\),then \(\frac{k}{n} \approx 0\)

If we take this case, \({e^k}\)is the approximate last factor, and then the approximation of \(n!\) and \(\left( {n - k} \right)!\) is depends upon\(\frac{{{n^k}}}{{k!}}\).

If\(\frac{n}{{n - k}} = 1\),then the product of k factors given by\(n! = {n^k}\)

Therefore, the approximation of \({P_{n,k}}\)is\({n^k}\).