A) \[6\]
B) \[5\]
C) \[4\]
D) \[3\]
Correct Answer: C
Solution :
\[\therefore \] Required number of ways \[{{=}^{2n+1}}{{C}_{2}}{{+}^{2n+1}}{{C}_{2}}+....{{+}^{2n+1}}{{C}_{n}}=255\] Now, \[{{(1+1)}^{2n+1}}{{=}^{2n+1}}{{C}_{0}}{{+}^{2n+1}}{{C}_{1}}+...{{+}^{2n+1}}{{C}_{n}}\] \[{{+}^{2n+1}}{{C}_{n+1}}{{+}^{2n+1}}{{C}_{n+2}}+...{{+}^{2n+1}}{{C}_{2n+1}}\] \[\Rightarrow \] \[{{2}^{2n+1}}{{=}^{2n+1}}{{C}_{0}}+{{2}^{(2n+1}}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+....\] \[{{+}^{2n+1}}{{C}_{n}}){{+}^{2n+1}}{{C}_{2n+1}}\] \[\Rightarrow \] \[{{2}^{2n+1}}=1+2,\,(255)+1\,\,(\because \,{{\,}^{n}}{{C}_{r}}{{=}^{n}}{{C}_{n-r}})\] \[\Rightarrow \] \[{{2}^{2n+1}}=2+510=512\] \[\Rightarrow \] \[{{2}^{2n+1}}={{2}^{9}}\,\,\Rightarrow \,\,2n+1=9\] \[\Rightarrow \] \[2n=8\Rightarrow n=4\]
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