Calculation of factorial numbers, and product in increasing order.

Question

Arrangement of following product in increasing order from left to right:

$A = 1000!$

$B = (400!)^2⋅(200!)$

$C = (500!)^2$

$D = (600!)⋅(300!)⋅(1000!)$

$E = (700!)⋅(300!)$

$\underline{\bf{My\; Try}}$: Here $(600!)\cdot(300!)\cdot(1000!)$ is largest, but I did not understand in which sequence the order is increase.

Help Required.

Thanks.

Answer 1

A useful lemma that you should prove:

Lemma: $(n - p)!\ p!\ \leq\ n!$ with equality only when $p = 0$ or $n$.

If you are having trouble with the proof of this fact, I have provided a hint below (put your cursor over the grey box).

Proof: Consider ${n \choose p}$.

This lemma tells us that $B, C, E < A$. Clearly $A < D$ so the chain ends with $A < D$. We just need to determine the relative sizes of $B, C,$ and $E$.

Note that $$\frac{B}{C} = \frac{(400!)^2 200!}{(500!)^2}=\frac{200!}{\left(\dfrac{500!}{400!}\right)^2}=\frac{200\times 199\times \dots\times 2\times 1}{(500\times 499\times\dots\times 402\times 401)^2} < 1$$ as each factor in the denominator is larger than each factor in the numerator, so $B < C$.

Similarly, one can compare $B$ and $E$ by considering $$\frac{B}{E} = \frac{(400!)^2 200!}{700!\ 300!} = \frac{\left(\dfrac{400!}{300!}\right)200!}{\dfrac{700!}{400!}}.$$ Likewise for $C$ and $E$ $$\frac{C}{E} = \frac{(500!)^2}{700!\ 300!} = \frac{\dfrac{500!}{300!}}{\dfrac{700!}{500!}}.$$

You will then be able to put the numbers $B, C,$ and $E$ is size order. Together with the facts that $B, C, E < A$ and $A < D$, you will have the complete order.