0

Prove that for every $n$,$i$ $\in$ $\mathbb{N}$ if $0$$\leq$$i$$\leq$$n$ then ${{n}\choose{i}}$$=$${{n}\choose{n-i}}$. Secondly, prove that if $i$$<$$j$$\leq$$n\over2$ then ${{n}\choose{i}}$$<$${{n}\choose{j}}$.

According to me, the first one is trivially true by expanding ${{n}\choose{n-i}}$ to get to ${{n}\choose{i}}$ (Is that correct though?). Concerning the second part, here's what I did: $($$i$$<$$j$$\leq$$n\over2$$)$ $\longrightarrow$ $($$-i$$>$$-j$$\geq$$-$$n\over2$$)$ $\longrightarrow$ $($$n-i$$>$$n-j$$\geq$$n$$-$$n\over2$$)$ $\longrightarrow$ $($$n-i$$>$$n-j$$\geq$$n\over2$$\geq$$j$$>$$i$$)$. This is where I got stuck when trying to prove:

$(n-i)$$!$$i$$!$ $>$ $(n-j)$$!$$j$$!$ since:

$n!\over{(n-i)!i!}$ $<$ $n!\over{(n-j)!j!}$ $\iff$ $(n-i)$$!$$i$$!$ $>$ $(n-j)$$!$$j$$!$

Slavik Egorov
  • 523

2 Answers2

1

Thus, we need to prove that: $$(n-i)(n-i-1)...(n-i-j+1)>(n-j)(n-j-1)...(n-j-i+1),$$

which is true because $n-i>n-j$.

Michael Rozenberg
  • 194,933
0

Let's say we proved this for a row $n$. Then, for $i\leq\tfrac n2$, we know $${n\choose i-2}<{n\choose i-1}<{n \choose i}$$ and so adding them gives us

$${n\choose i-2}+{n\choose i-1}<{n\choose i-1}+{n \choose i}$$ so that $${n+1\choose i-1}<{n+1\choose i}$$ which, by induction, proves that ${n+1\choose i}<{n+1\choose j}$ for $i<j\leq \frac n2$. This is not enough however, because we need the edge case $j=\frac{n+1}{2}$. Not to spoil everything for you, try to look at pascals triangle and see why the middle number is larger than its neighbours in even rows. Look at the row above it and try to express your ideas into mathematical reasonings.

  • I might be missing something, but why is ${{n}\choose{i-2}}$ $<$ ${{n}\choose{i-1}}$ $<$ ${{n}\choose{i}}$ correct? – Slavik Egorov Oct 05 '17 at 12:00
  • That's what you assume. I'm inexplicitly using induction here. Together with a base case, this is a solid proof. –  Oct 05 '17 at 12:28