The Factorial Pyramid

Relevant wiki: Factorials Problem Solving - Intermediate

The fraction in simplified form is :

$\dfrac{100\cancel{!}}{\cancel{99!}}\times\dfrac{98\cancel{!}}{\cancel{97!}}\times\cdots\times \dfrac{2!}{1!}\times\dfrac{0!}{2^{50}}$ $=\dfrac{\overbrace{100}^{2\times 50}\times \overbrace{98}^{2\times 49}\times \overbrace{96}^{2\times 48}\times\cdots\times \overbrace{2}^{2\times 1}}{2^{50}}$ $=\dfrac{\cancel{2^{50}}(50\times 49\times 48\times \cdots\times 1)}{\cancel{2^{50}}}$

$\large =\boxed{50!}$

First we can start by finding the bottom part of this gigantic numerator.

$\frac {1!}{0!}\quad =\quad 1$

$\frac {3!}{2!}\quad =\quad 3$

$\frac {4!}{3}\quad =\quad 4\times 2$

$\frac {5!}{4\times 2}\quad =\quad 5\times 3$

$\frac {6!}{5\times 3}\quad =\quad 6\times 4\times 2$

$\frac {7!}{6\times 4\times 2}\quad =\quad 7\times 5\times 3$

See the pattern here?

For all x! divided by the previous output, $\begin{cases} For\quad all\quad x\quad \in \quad the\quad set\quad of\quad all\quad even\quad numbers,\quad Output=x(x-2)(x-4)\cdots (4)(2) \\ For\quad all\quad x\quad \in \quad the\quad set\quad of\quad all\quad odd\quad numbers,\quad Output=x(x-2)(x-4)\cdots (5)(3) \end{cases}$

Let's plug in 100 for x into the first case, since 100 is an even number. We now have: $100\times 98\times 96\times 94\times 92\times 90\times \cdots \times 10\times 8\times 6\times 4\times 2$

We're not done here yet. We have to divide that whole thing by ${2}^{50}$ . Divide each term by 2, and we get: $50\times 49\times 48\times 47\times 46\times 45\times \cdots \times 5\times 4\times 3\times 2\times 1$

The problem tells you to put it into factorial form. In this case, we have $50!$ .

$\therefore$ The answer is $50$