Question

(n-3)! / (n-5)! = 6
n>=5
Vreau raspunsul cat mai clar daca se poate .

Answer 1

(n-3)! = (n-5)!(n-4)(n-3) => (n-3)!/(n-5)! = (n-3)(n-4)
n^2-7n+6=0 => solutiile sunt n1=6 si n2=1, dar n>=5 => singura solutie care convine este n =6

Answer 2

$\displaystyle \frac{(n-3)!}{(n-5)!} =6 ~~~~~~~~~~~~~~~~~~ n \geq 5 \Rightarrow n \in [5, \infty ) \\ \frac{(n-3)(n-4)(n-5)(n-6) \cdot ... \cdot 3 \cdot 2 \cdot 1}{(n-5)(n-6) \cdot ... \cdot 3 \cdot 2 \cdot 1} =6 \\ (n-3)(n-4)=6 \\ n^2-4n-3n+12=6 \\ n^2-7n+12=6 \\ n^2-7n=6-12 \\ n^2-7n=-6 \\ n^2-7n+6=0 \\ a=1,~b=-7,~c=6$
$\displaystyle \Delta=b^2-4ac=(-7)^2-4 \cdot 1 \cdot 6=49-24=25\ \textgreater \ 0 \\ n_1= \frac{7- \sqrt{25} }{2 \cdot 1}= \frac{7-5}{2}= \frac{2}{2} =1 \\ n_2= \frac{7+ \sqrt{25} }{2 \cdot 1} = \frac{7+5}{2} = \frac{12}{2} =6 \\ \left\begin{array}{ccc}n_1=1 \not \in [5, \infty)&&\\n_2=6 \in [5, \infty ) \\\end{array}\right\} \Rightarrow S=\{6\}$