A) 183
B) 186
C) 197
D) 190
E) 185
Correct Answer: E
Solution :
The number of required ways \[{{=}^{20}}{{C}_{2}}{{-}^{4}}{{C}_{2}}+1\] \[=\frac{20!}{2!18!}-\frac{4!}{2!2!}+1\] \[=\frac{20\times 19}{2}-\frac{4\times 3}{2}+1\] \[=190-6+1=185\]
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