Rút gọnb\(A=\frac{2-\mathrm{C}_{11}^{4}+\frac{11}{4} \mathrm{C}_{10}^{3}}{2 \mathrm{C}_{13}^{7}+\mathrm{C}_{12}^{7}-\mathrm{C}_{12}^{5}+\mathrm{C}_{14}^{7}}\) ta được
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Lời giải:
Báo saiáp dụng \(\mathrm{C}_{n}^{k}=\frac{n}{k} \cdot \mathrm{C}_{n-1}^{k-1}\)
Ta có:
\(\mathrm{C}_{11}^{4}=\frac{11}{4} \mathrm{C}_{10}^{3}, \mathrm{C}_{14}^{7}=\frac{14}{7} \mathrm{C}_{13}^{6}=2 \mathrm{C}_{13}^{6}\)
Mà \(\mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{n-k} \Rightarrow \mathrm{C}_{12}^{7}=\mathrm{C}_{12}^{12-7}=\mathrm{C}_{12}^{5}\)
\(A=\frac{2-\frac{11}{4} \mathrm{C}_{10}^{3}+\frac{11}{4} \mathrm{C}_{10}^{3}}{2 \mathrm{C}_{13}^{7}+\mathrm{C}_{12}^{7}-\mathrm{C}_{12}^{7}+2 \mathrm{C}_{13}^{6}}=\frac{2}{2\left(\mathrm{C}_{13}^{6}+\mathrm{C}_{13}^{7}\right)}=\frac{1}{\mathrm{C}_{14}^{7}}=\frac{1}{3432}\)