Definiție
Formule importante:
\[ C_n^k=C_n^{\,n-k},\quad \forall n,k\in\mathbb{N},\ 0\le k\le n,\ n\ge 1 \] formula combinărilor complementare
\[ C_n^k=C_{n-1}^k+C_{n-1}^{k-1},\quad \forall n,k\in\mathbb{N},\ 0\le k\le n,\ n\ge 1 \] formula de recurență pentru combinări
\[ C_{n+1}^{k+1}=\frac{n+1}{k+1}C_n^k \]
\[ C_n^{k+1}=\frac{n-k}{k+1}C_n^k \]
Exemple rezolvate
Exemplul 1
Rezolvare:
Patru băieți din cei 12 pot fi aleși în \(C_{12}^{4}\) moduri, iar două fete din cele 15 pot fi alese în \(C_{15}^{2}\) moduri.
Astfel, echipele pot fi formate în:
\[ C_{12}^{4}\cdot C_{15}^{2} = \frac{12!}{8!\cdot 4!}\cdot \frac{15!}{13!\cdot 2!} = \frac{8!\cdot 9\cdot 10\cdot 11\cdot 12}{8!\cdot 1\cdot 2\cdot 3\cdot 4} \cdot \frac{13!\cdot 14\cdot 15}{13!\cdot 1\cdot 2} = 495\cdot 105 = 51975 \]Prin urmare, se pot alcătui 51975 echipe.
Exemplul 2
Rezolvare:
a) \(\displaystyle C_9^4=\frac{9!}{4!\cdot(9-4)!}=\frac{9!}{4!\cdot5!}=\frac{5!\cdot6\cdot7\cdot8\cdot9}{4!\cdot5!}=\frac{6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4}=126\)
b) \(\displaystyle C_7^2=\frac{7!}{2!\cdot(7-2)!}=\frac{7!}{2!\cdot5!}=\frac{5!\cdot6\cdot7}{2!\cdot5!}=\frac{6\cdot7}{1\cdot2}=21,\quad C_7^1=\frac{7!}{1!\cdot(7-1)!}=\frac{7!}{1!\cdot6!}=\frac{6!\cdot7}{1!\cdot6!}=7\)
\(\displaystyle C_7^2-C_7^1=21-7=7(3-1)=14\)
c) \(\displaystyle \frac{C_{10}^5}{C_8^4}=\frac{\frac{10!}{5!\cdot5!}}{\frac{8!}{4!\cdot4!}}=\frac{10!}{5!\cdot5!}\cdot\frac{4!\cdot4!}{8!}=\frac{8!\cdot9\cdot10}{5!\cdot5!}\cdot\frac{4!\cdot4!}{8!}=\frac{9\cdot10\cdot4!\cdot4!}{5!\cdot5!}=\frac{9\cdot10}{5\cdot5}=\frac{18}{5}\)
Exemplul 3
Rezolvare:
a) \(\displaystyle C_n^3=\frac{n!}{3!\cdot(n-3)!}=\frac{(n-3)!\cdot(n-2)(n-1)n}{3!\cdot(n-3)!}=\frac{n(n-1)(n-2)}{1\cdot2\cdot3}\)
b) \(\displaystyle C_n^2=\frac{n!}{2!\cdot(n-2)!}=\frac{(n-2)!\cdot(n-1)n}{2!\cdot(n-2)!}=\frac{n(n-1)}{1\cdot2},\quad C_n^1=\frac{n!}{1!\cdot(n-1)!}=\frac{(n-1)!\cdot n}{1!\cdot(n-1)!}=n\)
\(\displaystyle C_n^2-C_n^1=\frac{n(n-1)}{2}-n=\frac{n(n-1)-2n}{2}=\frac{n(n-3)}{2}\)
c) \(\require{cancel}\displaystyle \frac{C_n^4-C_n^2}{C_n^3} =\frac{\frac{\cancel{n!}}{4!\cdot(n-4)!}-\frac{\cancel{n!}}{2!\cdot(n-2)!}}{\frac{\cancel{n!}}{3!\cdot(n-3)!}} =\frac{\frac{1}{4!\cdot(n-4)!}-\frac{1}{2!\cdot(n-2)!}}{\frac{1}{3!\cdot(n-3)!}}=\)
\(\displaystyle = \frac{\frac{1}{2!\cdot3\cdot4\cdot(n-4)!}-\frac{1}{2!\cdot(n-2)(n-3)(n-4)!}}{\frac{1}{2!\cdot3\cdot(n-3)(n-4)!}} = \frac{\frac{1}{\cancel{2!}\cdot3\cdot4\cdot\cancel{(n-4)!}}-\frac{1}{\cancel{2!}\cdot(n-2)(n-3)\cdot\cancel{(n-4)!}}}{\frac{1}{\cancel{2!}\cdot3\cdot(n-3)\cdot\cancel{(n-4)!}}}=\)
\(\displaystyle = \left(\frac{1}{12}-\frac{1}{(n-2)(n-3)}\right)\cdot 3(n-3) =\frac{(n-2)(n-3)-12}{12(n-2)(n-3)}\cdot 3(n-3) =\frac{(n-2)(n-3)-12}{4(n-2)} =\frac{n^2-5n-6}{4(n-2)} =\frac{(n-6)(n+1)}{4(n-2)}\)
Exemplul 4
Rezolvare:
C.E.: \(n\ge 2,\; n\in\mathbb{N}\)
\(\displaystyle C_n^2=\frac{n!}{2!\cdot(n-2)!}=\frac{(n-2)!\cdot(n-1)n}{2!\cdot(n-2)!}=\frac{n(n-1)}{2}\)
\(\displaystyle \frac{n(n-1)}{2}=6 \Rightarrow n(n-1)=12 \Rightarrow n^2-n-12=0\)
\(\displaystyle \Delta=49,\quad n_1=\frac{1-7}{2}=\frac{-6}{2}=-3\notin\mathbb{N}\)
\(\displaystyle n_2=\frac{1+7}{2}=\frac{8}{2}=4\in\mathbb{N}\)
\(\displaystyle n=4\)
