Ex1:
a) Pentru permutări (P) folosim formula:
[tex]P_{n} =1*2*3*...*n=n![/tex] (n! reprezintă n factorial)
Deci avem: [tex]P_{5} =1*2*3*4*5=120[/tex]
b) Pentru [tex]C^{k} _{n}[/tex] (combinări de n luate câte k) avem formula:
[tex]C^{k} _{n} =\frac{n!}{k!*(n-k)!} , 0\leq k\leq n[/tex]
Deci avem:
[tex]P_{3} =1*2*3=6[/tex]
[tex]C^{3} _{4} =\frac{4!}{3!*(4-3)!} =\frac{1*2*3*4}{1*2*3*1!} =\frac{4}{1} =4[/tex]
Răspuns: 6-4=2.
c) Pentru [tex]A^{k} _{n}[/tex] (aranjamente de n luate câte k) avem formula:
[tex]A^{k} _{n} =\frac{n!}{(n-k)!} , 0\leq k\leq n[/tex]
Deci avem:
[tex]P_{4} =1*2*3*4=24[/tex]
[tex]A^{2} _{5} =\frac{5!}{(5-2)!}=\frac{1*2*3*4*5}{1*2*3} =\frac{4*5}{1} =20[/tex]
Răspuns: 24-20=4
d) [tex]C^{0} _{6} -C^{1} _{6}+C^{2} _{6}-C^{3} _{6}=[/tex][tex]\frac{6!}{0!*(6-0)!} - \frac{6!}{1!*(6-1)!}+\frac{6!}{2!*(6-2)!}-\frac{6!}{3!*(6-3)!}=\frac{6!}{1*6!}-\frac{6!}{1*5!}+\frac{6!}{2*4!}-\frac{6!}{6*3!}[/tex]
0! este întotdeauna 1.
[tex]1-6+\frac{1*2*3*4*5*6}{2*1*2*3*4} -\frac{1*2*3*4*5*6}{6*1*2*3} = 1-6+\frac{5*3}{1}-\frac{4*5}{1} = 1-6+15-20=-10[/tex]
e) [tex]P_{2} =1*2=2[/tex]
[tex]C^{1} _{4} =\frac{4!}{1!*(4-1)!} =\frac{1*2*3*4}{1*3!} =\frac{4}{1} =4[/tex]
[tex]A^{1} _{3} =\frac{3!}{(3-1)!}=\frac{1*2*3}{2!} =\frac{3}{1} =3[/tex]
Răspuns: [tex]\frac{2+4}{3} =\frac{6}{3} =2[/tex]
f) [tex]0!+1!+2!+3!=1+1+2+6=10[/tex]
g) [tex]\frac{100!}{98!} -\frac{100!}{99!} =\frac{1*2*3*...*98*99*100}{1*2*3*...*98} -\frac{1*2*3*...*99*100}{1*2*3*...*99}=\frac{99*100}{1} -\frac{100}{1} =9900-100=9800[/tex]
h) [tex]\frac{(n+1)!}{(n-1)!} = \frac{1*2*3*...*(n-1)*n*(n+1)}{1*2*3*...*(n-1)}=n*(n+1)[/tex]
Ex 2:
a) [tex]\frac{(n+2)!}{n!}=12 <=> \frac{1*2*3*...*n*(n+1)*(n+2)}{1*2*3*...*n}=12 <=>\frac{(n+1)*(n+2)}{1}=12[/tex]
[tex](n+1)*(n+2)=12 <=> n^{2} +2*n+n+2=12 <=> n^{2} +3*n+2=12[/tex]
[tex]n^{2} +3*n-10=0\\a=1 \\b=3\\c=-10\\[/tex]
Δ=[tex]b^{2} -4*a*c=9+40=9[/tex]
[tex]n_{1,2} =\frac{-b+/-\sqrt{delta} }{2a} \\n_{1} =\frac{-3+7}{2} =\frac{4}{2} =2\\n_{2} = \frac{-3-7}{2} =\frac{-10}{2} =-5[/tex]
b) [tex]A^{1} _{n}+A^{2} _{n}=144<=>\frac{n!}{(n-1)!} +\frac{n!}{(n-2)!} =144 <=>\frac{1*2*3*...*(n-1)*n}{1*2*3*...*(n-1)}+ \frac{1*2*3*...*(n-2)*(n-1)*n}{1*2*3*...*(n-2)}=144[/tex][tex]\frac{n}{1} +\frac{n*(n-1)}{1} =144 <=>n+n^{2} -n=144<=>n^{2} =144 =>n=12[/tex]
c) [tex]9*A^{4} _{x}=A^{5} _{x} <=> 9*\frac{x!}{(x-4)!} =\frac{x!}{(x-5)!}[/tex]
Se împarte relația la [tex]\frac{x!}{(x-4)!}[/tex] și se obține:
[tex]\frac{x!}{(x-5)!}:\frac{x!}{(x-4)!}=9 <=>\frac{x!}{(x-5)!}*\frac{(x-4)!}{x!}=9[/tex]
x! se simplifică, iar asemenea celorlalor exerciții, (x-4)! și (x-5)! au în comun (x-4)! și se simplifică, deci vom rămâne cu:
[tex]\frac{1*1}{x-5} =9 <=>9*x-45=1<=>9*x=46[/tex]
