A) \[\left( \begin{matrix} 30 \\ 10 \\ \end{matrix} \right)\]
B) \[\left( \begin{matrix} 30 \\ 15 \\ \end{matrix} \right)\]
C) \[\left( \begin{matrix} 60 \\ 30 \\ \end{matrix} \right)\]
D) \[\left( \begin{matrix} 31 \\ 10 \\ \end{matrix} \right)\]
Correct Answer: A
Solution :
| [a] To find |
| \[^{30}{{C}_{0}}{{\,}^{30}}{{C}_{10}}{{-}^{30}}{{C}_{1}}{{\,}^{30}}{{C}_{11}}+{{\,}^{30}}{{C}_{2}}{{\,}^{30}}{{C}_{12}}-...\] |
| \[+{{\,}^{30}}{{C}_{20}}{{\,}^{30}}{{C}_{30}}\] |
| We know that \[{{(1+x)}^{30}}={{\,}^{30}}{{C}_{0}}+{{\,}^{30}}{{C}_{1}}x+{{\,}^{30}}{{C}_{2}}{{x}^{2}}\] |
| \[+.....+{{\,}^{30}}{{C}_{20}}{{x}^{20}}+.....{{\,}^{30}}{{C}_{30}}{{x}^{30}}...(1)\] |
| \[{{(x-1)}^{30}}={{\,}^{30}}{{C}_{0}}{{x}^{30}}-{{\,}^{30}}{{C}_{1}}{{x}^{29}}+....+{{\,}^{30}}{{C}_{10}}{{x}^{20}}\] |
| \[{{-}^{30}}{{C}_{11}}{{x}^{19}}+{{\,}^{30}}{{C}_{12}}{{x}^{18}}+...{{\,}^{30}}{{C}_{30}}{{x}^{0}}...(2)\] |
| Multiplying \[e{{q}^{n}}\] (1) and (2), we get |
| \[{{({{x}^{2}}-1)}^{30}}=(\,1\,)\,\,\times \,\,(\,2\,)\] |
| Equating the coefficients of \[{{x}^{20}}\] on both sides, |
| we get |
| \[^{30}{{C}_{10}}={{\,}^{30}}{{C}_{0}}{{\,}^{30}}{{C}_{10}}-{{\,}^{30}}{{C}_{1}}{{\,}^{30}}{{C}_{11}}+\] |
| \[^{30}{{C}_{2}}{{\,}^{30}}{{C}_{12}}-......+{{\,}^{30}}{{C}_{20}}{{\,}^{30}}{{C}_{30}}\] |
| \[\therefore \] Req. value is \[^{30}{{C}_{10}}\] |
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