Simply calculate $\frac{ \frac{65536!}{(65536 - 100)! }}{65536^{100}}$

Question

I want to calculate this value exactly: $C=\frac{ \frac{65536!}{(65536 - 100)! }}{65536^{100}}$. Or at least I want to obtain an a convincing lower bound to argue that this value is close to 1.

It can be easily seen that the numerator is a permutation value, aka $P^{65536}_{100}$, also note that $65536=2^{16}$, if this may help you simplfy the expression.

One can simply lower bound this value $C\geq \frac{ (65536-99)^{99} }{65536^{99}} \approx 86\% $. Could anyone offer me a tigher bound? Or, if you can do the best, obtain the exact value? Thanks a lot!

Answer 1

Consider the more general case of $$A=\frac{\frac{a!}{(a-b)!} }{a^b}$$ Take logatihms and use Stirling approximation for large $a$.

You should obtain $$\log(A)=-\frac{(b-1) b}{2 a}-\frac{(b-1) b (2 b-1)}{12 a^2}-\frac{(b-1)^2 b^2}{12 a^3}+O\left(\frac{1}{a^4}\right)$$

Using your numbers and my phone $$\log\Bigg(\frac{ \frac{65536!}{(65536 - 100)! }}{65536^{100}}\Bigg)=-\frac{5317713913875}{70368744177664}$$

$$\frac{30}{397}<\frac{5317713913875}{70368744177664} <\frac{73}{966}$$ giving $$ 0.92721542<A<0.92721784$$

Answer 2

Let’s be brave and expand the numerator as shown below. $$C=\dfrac{\dfrac{65536!}{(65536 - 100)! }}{65536^{100}}=\dfrac{\dfrac{65536\times 65535\times\dots\times 65437\times 65436\times \dots\times 2\times 1}{65436\times 65435\times 65434\times\dots\times 2\times 1}}{65536^{100}}$$

When we simplify the above expression on the right hand side, we get, $$C=\dfrac{65536\times 65535\times\dots\times 65438\times 65437}{65536^{100}}.$$

Note that the numerator and denominator on the right hand side has equal number of multiplicands. Therefore, we shall write, $$C=1\times\dfrac{65535}{65536}\times\dfrac{65534}{65536}\times\dots\times\dfrac{65438}{65536}\times\dfrac{65437}{65536}.$$

This can be written as, $$C=\left(1-\dfrac{1}{2^{16}}\right)\times\left(1-\dfrac{2}{2^{16}}\right)\times\dots\times\left(1-\dfrac{50}{2^{16}}\right)\times\dots\times\left(1-\dfrac{98}{2^{16}}\right)\times\left(1-\dfrac{99}{2^{16}}\right).$$

Now, we are in a position to use a trick to further simplify the expression on the right hand side. Let’s take the first and the last multiplicands. $$\left(1-\dfrac{1}{2^{16}}\right)\times\left(1-\dfrac{99}{2^{16}}\right)=1+\dfrac{99}{2^{32}}-\dfrac{100}{2^{16}}$$

The multiplication between the second and the penultimate multiplicands gives, $$\left(1-\dfrac{2}{2^{16}}\right)\times\left(1-\dfrac{98}{2^{16}}\right)=1+\dfrac{196}{2^{32}}-\dfrac{100}{2^{16}}.$$

There are 49 such multiplications and the $49^{\text{th}}$ multiplication gives, $$\left(1-\dfrac{2}{49^{16}}\right)\times\left(1-\dfrac{51}{2^{16}}\right)=1+\dfrac{2499}{2^{32}}-\dfrac{100}{2^{16}}.$$

Therefore, we can write down the expresssion for $C$ in the following form. $$C=\left(1+\dfrac{99}{2^{32}}-\dfrac{100}{2^{16}}\right)\times\left(1+\dfrac{196}{2^{32}}-\dfrac{100}{2^{16}}\right)\times\dots\times\left(1+\dfrac{2499}{2^{32}}-\dfrac{100}{2^{16}}\right)\times\left(1-\dfrac{50}{2^{16}}\right)$$

Now we can determine the upper and the lower bounds of the given expression. $$\left(1+\dfrac{2499}{2^{32}}-\dfrac{100}{2^{16}}\right)^{49}\times\left(1-\dfrac{50}{2^{16}}\right)\gt C\gt\left(1+\dfrac{99}{2^{32}}-\dfrac{100}{2^{16}}\right)^{49}\times\left(1-\dfrac{50}{2^{16}}\right)$$

When the two bounds are simplified, this looks like, $$\small{0.927224239379141704120285686818\gt\dfrac{\dfrac{65536!}{(65536 - 100)! }}{65536^{100}}\gt0.9271988127178732384044710631461}.$$