Rozwiąż nierównośc

[damian18833]

a) \(\displaystyle{ {n \choose 4}}\)>\(\displaystyle{ {n \choose 5}}\)

b) \(\displaystyle{ {n \choose n-4}}\)<\(\displaystyle{ {n \choose n-3}}\)

[ares41]

\(\displaystyle{ {n \choose 4}>{n \choose 5}\\ \frac{n!}{(n-4)! \cdot 4!} > \frac{n!}{(n-5)! \cdot 5!}\\n! \cdot (n-5)! \cdot 5!>n! \cdot (n-4)! \cdot 4!\\(n-5)! \cdot 5!> (n-4)! \cdot 4!\\(n-5)! \cdot 5> (n-4)!\\(n-5)! \cdot 5> (n-4) \cdot (n-5)!\\5>n-4\\n<9}\)