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I am struggling some math problems.

Fighting some problems, I find out a rule.

$$$$

Could you please see the table below?

enter image description here

HERE is my question!

I (may) found out the inequality

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$$\binom {2n}{k} + \binom {2n}{n-k} \ < \ \binom{2n}{n} + \binom{2n}{0} \ \ \text{for} \ \ k<n \ \ \cdots(1)$$

or

$$\binom {2n}{k} + \binom {2n}{n-k} \ < \ \binom{2n}{l} + \binom{2n}{n-l} \ \ \text{for} \ \ k<l \le n \ \ \cdots(2)$$

$$\text{Actually (2) implies (1)}$$ $$$$

  1. Is these inequality exist already?
  2. Is these inequality true, INTUITIVELY?
  3. Is these inequality TRUE?
  4. If these are right, then how can I proof?

$$$$ (I wanna believe these are true, and then these will proof simply...)

Thank you for your attention to this matter.

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  • $\begingroup$ It fails for $k=0$. $\endgroup$ – lhf Oct 6 '18 at 16:19
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    $\begingroup$ The inequality $(1)$ does not hold for $(n,k)=(2,1),(3,1),(3,2)$. $\endgroup$ – mathlove Oct 7 '18 at 3:52

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