$\begin{array}{1 1}(A)\;55\\(B)\;66\\(C)\;77\\(D)\;88\end{array} $

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- $n!=n(n-1)(n-2)(n-3).....(1)(2)(3)$

There are two possible cases.

Case I:Five 1's one 2's one 3's

Number of numbers =$\large\frac{7!}{5!}$

$\Rightarrow \large\frac{7\times 6\times 5!}{5!}$$=42$

Case II: Four 1's three 2's

Number of numbers =$\large\frac{7!}{4!3!}$

$\Rightarrow \large\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2}$$=35$

Total number of numbers =42+35=77

Hence (C) is the correct answer.

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