A) \[{{n}^{2}}+n-110=0\]
B) \[{{n}^{2}}+5n-84=0\]
C) \[{{n}^{2}}+3n-180=0\]
D) \[{{n}^{2}}+2n-80=0\]
Correct Answer: C
Solution :
\[\frac{^{n+2}{{C}_{6}}}{^{n-2}{{P}_{2}}}=11\]\[\Rightarrow \] \[\frac{(n+2)!}{6!(n-4)!}=11.\frac{(n-2)!}{(n-4)!}\] \[\Rightarrow \,\,\,\,(n+2)!=11.\] \[6!(n-2)!\] \[\Rightarrow \,\,\,(n+2)\,\,(n+1)\,\,n(n-1)=11.6!\] \[\Rightarrow \,\,\,(n+2)\,\,(n+1)\,\,n\,(n-1)=11.10.9.8\] \[n+2=11\] \[\Rightarrow \,\,n=9\] Which satifies the \[{{n}^{2}}+3n-108=0\]
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