A) \[1\]
B) \[0\]
C) \[11\]
D) \[10\]
Correct Answer: A
Solution :
Given, \[^{n}{{C}_{10}}{{=}^{n}}{{C}_{11}}\] \[\Rightarrow \] \[\frac{n!}{10!(n-10)!}=\frac{n!}{11!(n-11)!}\] \[\Rightarrow \] \[11\times 10!(n-11)!=0!(n-10)(n-11)!\] \[\Rightarrow \] \[11=n-10\] \[\Rightarrow \] \[n=11+10=21\] Now, \[^{n}{{C}_{21}}{{=}^{21}}{{C}_{21}}=1\]
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