A) 1
B) 20
C) 210
D) 1540
Correct Answer: B
Solution :
Given, \[^{n}{{C}_{12}}{{=}^{n}}{{C}_{8}}\] \[\Rightarrow \] \[\frac{n!}{(n-12)!12!}=\frac{nQ}{(n-8)!8!}\] \[\Rightarrow \] \[(n-8)(n-9)(n-10)(n-11)\] \[=12\times 11\times 10\times 9\] \[\Rightarrow \] \[n-8=12\Rightarrow n=20\] \[\therefore \] \[^{20}{{C}_{19}}=20\]
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