Aranjamente
Numărul submulțimilor ordonate cu \(k\) elemente ale unei mulțimi cu \(n\) elemente se notează \(A_n^k\), se citește „aranjamente de \(n\) luate câte \(k\)” și este:
\[ A_n^k = \frac{n!}{(n-k)!} = n(n-1)(n-2)\ldots(n-k+1), \quad \forall n \in \mathbb{N},\; k \le n,\; k \in \mathbb{N}. \]
Exemple rezolvate
Să se calculeze: a) \(A_7^2\); b) \(A_9^3\); c) \(A_{10}^2\); d) \(A_6^3 + A_6^2\); e) \(A_8^2 \cdot A_7^3\); f) \(\frac{A_6^4}{A_6^2}\)
Rezolvare:
a) \(A_7^2 = \frac{7!}{(7-2)!} = \frac{7!}{5!} = \frac{7 \cdot 6 \cdot 5!}{5!} = 7 \cdot 6 = 42\)
b) \(A_9^3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} = \frac{9 \cdot 8 \cdot 7 \cdot 6!}{6!} = 9 \cdot 8 \cdot 7 = 504\)
c) \(A_{10}^2 = \frac{10!}{(10-2)!} = \frac{10!}{8!} = \frac{10 \cdot 9 \cdot 8!}{8!} = 10 \cdot 9 = 90\)
d) \(A_6^3 + A_6^2 = \frac{6!}{3!} + \frac{6!}{4!} = \frac{6 \cdot 5 \cdot 4 \cdot 3!}{3!} + \frac{6 \cdot 5 \cdot 4!}{4!} = 6 \cdot 5 \cdot 4 + 6 \cdot 5 = 120 + 30 = 150\)
e) \(A_8^2 \cdot A_7^3 = \frac{8!}{6!} \cdot \frac{7!}{4!} = \frac{8 \cdot 7 \cdot 6!}{6!} \cdot \frac{7 \cdot 6 \cdot 5 \cdot 4!}{4!} = (8 \cdot 7)(7 \cdot 6 \cdot 5) = 56 \cdot 210 = 11760\)
f) \(\frac{A_6^4}{A_6^2} = \frac{\frac{6!}{2!}}{\frac{6!}{4!}} = \frac{6!}{2!} \cdot \frac{4!}{6!} = \frac{4!}{2!} = \frac{4 \cdot 3 \cdot 2!}{2!} = 4 \cdot 3 = 12\)
Calculați: a) \(\frac{A_8^5 + A_8^4}{A_8^3}\); b) \(A_9^3 – P_3\); c) \(\frac{A_7^3 + A_7^2}{A_7^2}\); d) \(A_{10}^2 + A_{10}^3 + A_{10}^4\); e) \(\frac{A_{11}^7 + A_{11}^6}{A_{11}^5 + A_{11}^4}\)
Rezolvare:
a) \(\frac{A_8^5 + A_8^4}{A_8^3} = \frac{\frac{8!}{3!}+\frac{8!}{4!}}{\frac{8!}{5!}} = \frac{\frac{8!}{3!}+\frac{8!}{3!\cdot4}}{\frac{8!}{5!}} = \frac{\frac{8!}{3!}\left(1+\frac14\right)}{\frac{8!}{5!}} = \frac{\frac{8!}{3!}\cdot\frac54}{\frac{8!}{5!}} = \frac{8!}{3!}\cdot\frac54\cdot\frac{5!}{8!} = \frac54\cdot\frac{5!}{3!} = \frac54\cdot\frac{3!\cdot4\cdot5}{3!} = \frac54\cdot4\cdot5 = 25\)
b) \(A_9^3 – P_3 = \frac{9!}{6!} – 3! = \frac{6!\cdot7\cdot8\cdot9}{6!} – 6 = 7\cdot8\cdot9 – 6 = 504 – 6 = 498\)
c) \(\frac{A_7^3 + A_7^2}{A_7^2} = \frac{\frac{7!}{4!}+\frac{7!}{5!}}{\frac{7!}{5!}} = \frac{\frac{7!}{4!}+\frac{7!}{4!\cdot5}}{\frac{7!}{5!}} = \frac{\frac{7!}{4!}\left(1+\frac15\right)}{\frac{7!}{5!}} = \frac{\frac{7!}{4!}\cdot\frac65}{\frac{7!}{5!}} = \frac{7!}{4!}\cdot\frac65\cdot\frac{5!}{7!} = \frac65\cdot\frac{5!}{4!} = \frac65\cdot\frac{4!\cdot5}{4!} = 6\)
d) \(A_{10}^2 + A_{10}^3 + A_{10}^4 = \frac{10!}{8!} + \frac{10!}{7!} + \frac{10!}{6!} = \frac{10!}{8!} + \frac{10!}{8!}\cdot8 + \frac{10!}{8!}\cdot8\cdot7 = \frac{10!}{8!}(1+8+56) = \frac{10!}{8!}\cdot65 = 10\cdot9\cdot65 = 5850\)
e) \(\frac{A_{11}^7 + A_{11}^6}{A_{11}^5 + A_{11}^4} = \frac{\frac{11!}{4!}+\frac{11!}{5!}}{\frac{11!}{6!}+\frac{11!}{7!}} = \frac{\frac{11!}{4!}+\frac{11!}{4!\cdot5}}{\frac{11!}{6!}+\frac{11!}{6!\cdot7}} = \frac{\frac{11!}{4!}\left(1+\frac{1}{5}\right)}{\frac{11!}{6!}\left(1+\frac{1}{7}\right)} = \frac{\frac{11!}{4!}\cdot\frac{6}{5}}{\frac{11!}{6!}\cdot\frac{8}{7}} = \frac{11!}{4!}\cdot\frac{6}{5}\cdot\frac{6!}{11!}\cdot\frac{7}{8} = \frac{6}{5}\cdot\frac{6!}{4!}\cdot\frac{7}{8} = \frac{6}{5}\cdot(6\cdot5)\cdot\frac{7}{8} = \frac{6\cdot30\cdot7}{5\cdot8} = \frac{252}{8} = 31,5\)
Rezolvare:
a) \(\frac{A_n^2}{A_n^1} = \frac{\frac{n!}{(n-2)!}}{\frac{n!}{(n-1)!}} = \frac{n!}{(n-2)!}\cdot\frac{(n-1)!}{n!} = \frac{(n-1)!}{(n-2)!} = \frac{(n-2)!\cdot(n-1)}{(n-2)!} = n-1\)
b) \(\frac{A_n^3 – A_n^2}{A_n^2} = \frac{\frac{n!}{(n-3)!}-\frac{n!}{(n-2)!}}{\frac{n!}{(n-2)!}} = \frac{\frac{n!}{(n-3)!}-\frac{n!}{(n-3)!\cdot(n-2)}}{\frac{n!}{(n-2)!}} = \frac{\frac{n!}{(n-3)!}\left(1-\frac{1}{n-2}\right)}{\frac{n!}{(n-2)!}} = \frac{\frac{n!}{(n-3)!}\cdot\frac{n-3}{n-2}}{\frac{n!}{(n-2)!}} = \frac{n!}{(n-3)!}\cdot\frac{n-3}{n-2}\cdot\frac{(n-2)!}{n!} = \frac{(n-2)!}{(n-3)!}\cdot\frac{n-3}{n-2} = (n-2)\cdot\frac{n-3}{n-2} = n-3\)
c) \(\frac{A_n^4 – A_n^3}{A_n^3 – A_n^2} = \frac{\frac{n!}{(n-4)!}-\frac{n!}{(n-3)!}}{\frac{n!}{(n-3)!}-\frac{n!}{(n-2)!}} = \frac{\frac{n!}{(n-4)!}-\frac{n!}{(n-4)!\cdot(n-3)}}{\frac{n!}{(n-3)!}-\frac{n!}{(n-3)!\cdot(n-2)}} = \frac{\frac{n!}{(n-4)!}\left(1-\frac{1}{n-3}\right)}{\frac{n!}{(n-3)!}\left(1-\frac{1}{n-2}\right)} = \frac{\frac{n!}{(n-4)!}\cdot\frac{n-4}{n-3}}{\frac{n!}{(n-3)!}\cdot\frac{n-3}{n-2}} = \frac{n!}{(n-4)!}\cdot\frac{n-4}{n-3}\cdot\frac{(n-3)!}{n!}\cdot\frac{n-2}{n-3} = \frac{(n-3)!}{(n-4)!}\cdot\frac{n-4}{n-3}\cdot\frac{n-2}{n-3} = (n-3)\cdot\frac{n-4}{n-3}\cdot\frac{n-2}{n-3} = \frac{(n-4)(n-2)}{n-3}\)
Rezolvare:
C.E.: \(n \in \mathbb{N},\; n \ge 2\)
\(A_n^2=20\)
\(\frac{n!}{(n-2)!}=20 \;\Rightarrow\; \frac{(n-2)!\cdot (n-1)\cdot n}{(n-2)!}=20\)
\(n(n-1)=20 \;\Rightarrow\; n^2-n-20=0 \;\Rightarrow\; (n-5)(n+4)=0\)
\(n=5 \;\text{sau}\; n=-4\)
Verificare:
\(n=5 \Rightarrow A_5^2=\frac{5!}{3!}=\frac{3!\cdot4\cdot5}{3!}=20\)
\(n=-4 \notin \mathbb{N}\)
