Arrangement of $100$ articles numbered $n_{1},n_{2},n_{3},n_{4} \cdots n_{100}$. There are $100$ articles numbered $n_{1},n_{2},n_{3},n_{4} \cdots n_{100}$
  They are arranged in all possible ways. How many arrangments would be there 
  in which $n_{28}$ will always be before $n_{29}$ 


$a.)\ 5050\times 99! \\
b.)\ 5050\times 98! \\
\color{green}{c.)\ 4950\times 98!} \\
d.)\ 4950\times 99! \\ $

I integrated $n_{28}$ and $n_{29}$ in one term and concluded 
that answer will be $99!$. 

But that's not in options.

I look for a short and simple way.

I have studied maths upto $12$th grade.
Intuitive and Simple

Use the symmetry of the problem. Half of the cases of all arrangements, which is half of $100!$ should be the answer. This can be written as $\frac{1}{2}\times 99\times100\times 98!$ which is $4950 \times 98!$
Of the total arrangements, half of them have n28 before n29 and other half arrangements have n28 after n29.
So,
  total arrangements =100!
  half of them = 100!/2=50*99!=50*99*98!=4950*98!
hope you understood!