Question

ajutati-ma   va rog  sa rezolv inecuatiile pe tema"Elemente de combinatorica "

Answer 1

 $3a) \\ \frac{n!}{(n-2)!}=12 \\ \\ \frac{1*2*3*....(n-2)*(n-1)*n}{1*2*3*....(n-2)} =12 \\ \\ (n-1)*n = 12 \\ \\ n^{2} -n - 12 = 0 \\ \\ n_1= \frac{1+ \sqrt{1+4*12} }{2}= \frac{1+ \sqrt{49} }{2} = \frac{1+7}{2} = \frac{8}{2} = 4 \\ \\ n_2 <0\, \, se\, \, exclude \\ \\ =>n=4$


$3b) \\ \frac{n!}{(n-4)!}= \frac{x}{y} \frac{22n!}{(n-3)!} \\ \\ \frac{(n-3)!}{(n-4)!}= \frac{22n!}{n!} \\ \\ \frac{(n-3)!}{(n-4)!}= 22\frac{n!}{n!} \\ \\ \frac{1*2*3*....(n-4)*(n-3)}{1*2*3*....(n-4)} = 22\frac{n!} {n!} \\ \\ n-3 = 22 \\ \\ n=22+3 \\ \\ n=25$


$3c) \\ \frac{n!}{(n-5)!}= \frac{6n!}{(n-3)!} \\ \\ \frac{(n-3)!}{n-5)!} = \frac{6n!}{n!} \\ \\ (n-4)*(n-3)=6 \\ \\ n^{2}-7n+12 = 6 \\ \\ n^{2}-7n+6=0 \\ \\ n_1= \frac{7+ \sqrt{49-4*6} }{2}=\frac{7+ \sqrt{25} }{2} = \frac{7+5}{2}= \frac{12}{2}=6 \\ \\ n_2 = \frac{7-5}{2} = \frac{2}{2}= 1 \, \, se\, \, exclude \\ \\ => n = 6$