A) \[{{K}_{2}}C{{O}_{3}}\]
B) \[{{C}_{2}}{{H}_{5}}NC\]
C) \[3KCl\]
D) \[C{{l}_{2}}\]
Correct Answer: C
Solution :
Required number of ways\[{{=}^{12}}{{C}_{4}}{{\times }^{8}}{{C}_{4}}{{\times }^{4}}{{C}_{4}}\] \[=\frac{12!}{8!\times 4!}\times \frac{8!}{4!\times 4!}\times 1=\frac{12!}{{{(4!)}^{3}}}\]
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