Prove that $$87!<16! \left(52^{71}\right)$$ I do not how can i compare between the factorials or what the procedure to solve such questions?

  • 3
    $\begingroup$ $87!=16! \times 17 \times \dots \times 87$ $\endgroup$
    – Naj Kamp
    Jan 27, 2018 at 13:02

3 Answers 3


You know that $87!=16! \cdot 17 \cdot 18 \dots 87$, hence you want to show that

$$17 \cdot 18\cdot 19 \dots 87 < 52^{71}$$

Now notice that on the left hand side you have $71$ terms with $52$ being the middle one. Hence you can group that in pairs (first with last, second with second last etc. and finally leave $52$ alone). You can say the following:


Hence the product in each pair is less than $52^2$

Do this for all $x$'s between $1$ and $35$ to get that the product is less than $52^{71}$ since you have $35$ pairs for which it holds and then you add the $52$ factor



Note that

$$87!<16! \left(52^{71}\right)\iff \frac{87!}{16!}<52^{71}\iff\binom{87}{71}<\frac{52^{71}}{71!}$$

which is true from $\ { n \choose k} \leq \frac{n^k}{k!}. ( \, 1-\frac {k}{2n} ) \,^{k-1} $ since

$$\binom{87}{71}\le \frac{87^{71}}{71!}\cdot\left(1-\frac{71}{2\cdot 87}\right)^{70}=\frac{52^{71}}{71!}\cdot \frac{87^{71}}{52^{71}}\cdot \frac{103^{71}}{174^{71}}\cdot\frac{174}{103}=\frac{174}{103}\cdot\left(\frac{87\cdot103}{52\cdot 174}\right)^{71}\cdot \frac{52^{71}}{71!}$$

$$=\frac{174}{103}\cdot \left(\frac{103}{104}\right)^{71}\cdot \frac{52^{71}}{71!}<0.851\cdot \frac{52^{71}}{71!}$$


AM-GM gives: $$\frac1{71}\sum_{i=1}^{71}(i+16)>\sqrt[71]{\prod_{i=1}^{71}(i+16)}$$ So: $$\frac1{71}\left(\frac{71(71+1)}{2}+71\cdot 16\right)>\sqrt[71]{\frac{87!}{16!}}$$ or: $$52^{71}>\frac{87!}{16!}\implies 16!(52^{71})>87!$$

  • $\begingroup$ That is great solution ,thank you @Mastrem $\endgroup$ Jan 27, 2018 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.