Question

Помогите пожалуйста))))))
№ 2,3,4

Answer 1

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

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

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

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

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