KOMBINATORIKA

Určete, kolikati nulami bude končit číslo 80!

Určete, kolikati nulami bude končit číslo 130!

$${{{1}\over{2!}} + {{1}\over{3!}}}$$

$${{{6! + 4! - 5!}\over{3!}}}$$

$${{{8! + 7! - 6!}\over{9! - 8!}}}$$

$${{{10!}\over{3!\cdot 4!\cdot 5!}}}$$

$${{{(3!)^2}\over{(2^2)!}}}$$

$${{{(n + 1)!}\over{n!}} + {{n!}\over{(n - 2)!}}}$$

Určete, kolikati nulami bude končit číslo 130!

$${{{1}\over{2!}} + {{1}\over{3!}}}$$

$${{{6! + 4! - 5!}\over{3!}}}$$

Dokažte, že platí: $${{(n + 1)}^2\cdot{(n!)}^2 - [(n + 1)!]^2 = 0}$$

Dokažte, že platí: $${(n + 2)! + (n + 3)! + (n + 4)! = (n + 2)!\cdot(n + 4)^2}$$

$${{{(n + 1)!}\over{n!}} - 5\cdot{{(n + 3)!}\over{(n + 1)!}} + {{3{n}!}\over{(n - 2)!}}}$$

Řešte soustavu rovnic: $${\binom{x}{y} = \binom{x}{y + 1} }$$ $${\frac{x!}{(x - 2)!} = 20 }$$

$${\binom{x - 1}{x - 2} + \binom{x - 2}{x - 4} = 4}$$

$${\binom{x}{x - 1} + \binom{x}{x - 2} = \frac{1}{2} \cdot (x^2 + 1)}$$

$${\binom{n + 3}{n + 1} = 7 + 3 \cdot \binom{n - 1}{n - 2} }$$