Question

Помогите пожалуйста))))))
№ 2,3,4

Answer 1

$3!=1*2*3=6\; ,\; \; \; 5!=1*2*3*4*5=120\; ,\\\\ 8!=1*2*3*4*5*6*7*8=5!*6*7*8=120*6*7*8=40320\\\\ \frac{10!}{6!2!}= \frac{6!*7*8*9*10}{6!2!}=7*8*9*5=2520\\\\ \frac{9!4!}{16!} = \frac{9!4!}{9!*10*11*12*13*14*15*16} = \frac{1*2*3*4}{10*11*12*13*14*15*16} =\\\\ =\frac{1}{10*11*2*13*14*15*4} = \frac{1}{4586400}$

$3)\; \; \frac{n!}{(n-2)!} =12\; ,\; \; \frac{(n-2)!*(n-1)*n}{(n-2)!}=12\; ,\; \; (n-1)n=12\\\\n^2-n-12=0\; ,\; \; n_1=-3,\; n_2=4 \\\\b)\; \; \frac{n!}{(n-4)!} = \frac{22n!}{(n-3)!}\; ,\; \; \; \; (n-3)(n-2)(n-1)n=22(n-2)(n-1)n\\\\n-3=22\; ,\; \; n=25\\\\c)\; \; \frac{n!}{(n-5)!} = \frac{6n!}{(n-3)!} \\\\ \frac{1}{(n-5)!} = \frac{6}{(n-3)!} \; ,\; \; \; \frac{(n-3)!}{(n-5)!} =6\; ,\; \; \frac{(n-5)!(n-4)(n-3)}{(n-5)!} =6\\\\(n-4)(n-3)=6\; ,\; \; n^2-7n+6=0\\\\n_1=1,\; n_2=6$

$4)\; \; \frac{(n-1)!}{(n-3)!} \leq 20\; ,\; \; \frac{(n-3)!(n-2)(n-1)}{(n-3)!} \leq 20\\\\(n-2)(n-1) \leq 20\\\\n^2-3n-18 \leq 0\\\\n_1=-3\; ,\; n_2=6\\\\+++(-3)---(6)+++\\\\n\in [-3,6\, ]\\\\n\in N\; \to \; n=1,2,3,4,5,6$

$b)\; \; \frac{16n!}{(n-1)!} \ \textgreater \ \frac{5n!}{(n-2)!} \\\\ \frac{16(n-2)!}{(n-1)!} \ \textgreater \ 5\; ,\; \; \frac{16}{n-1} \ \textgreater \ 5\; ,\; \; \frac{16-5(n-1)}{n-1}\ \textgreater \ 0\; ,\; \; \frac{21-5n}{n-1} \ \textgreater \ 0\; ,\; \; \frac{5n-21}{n-1}\ \textless \ 0\\\\+++(1)---(\frac{21}{5}=4,2)+++\\\\n\in (1,\frac{21}{5})\\\\n\in N\; \to \; \; n=1,2,3,4$

$c)\; \; \frac{(n-4)!}{(n-2)!} \geq \frac{1}{20} \\\\ \frac{1}{(n-3)(n-2)} \geq \frac{1}{20} \; ,\; \; \frac{1}{(n-3)(n-2)} -\frac{1}{20} \geq 0\\\\ \frac{20-(n-3)(n-2)}{20(n-3)(n-2)} \geq 0\\\\ \frac{20-n^2+5n-6}{20(n-3)(n-2)} \geq 0\\\\ \frac{n^2-5n-14}{20(n-3)(n-2)} \geq 0\\\\n^2-5n-14\ \textgreater \ 0\; \; pri\; \; n\in R\; ,\; t.k.\; D\ \textless \ 0\\\\(n-3)(n-2)\ \textgreater \ 0\\\\+++(2)---(3)+++\\\\n\in (-\infty ,2)U(3,+\infty )\\\\n\in N\; \to \; n=1,4,5,6,7,...\\\\oo$