# 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.

• "Always" is weird. Always how? For any particular arrangement, either $n_{28}$ is before or after $n_{29}$. Oct 5 '15 at 14:28
• @ThomasAndrews: you mean the word always is redundant in the question.
– R K
Oct 5 '15 at 14:30
• It is redundant in a way that makes it sound like you might mean something else, yes. You are seeking the number of all permutations with the condition that article 28 comes before 29. But the always seems to apply to the condition on a single permutation. Oct 5 '15 at 14:33

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 $4950 \times 98!$
Half of the $100\times99 = 4950$ ways the two specified numbers can be placed satisfy the stipulations, and the rest can be permuted in $98!$ ways
One more approach: once you have $k$ slot fixed for $n_{28}$, you have $n-k$ slots in which you can put $n_{29}$. Since you don't care about the remaining items, you have $98!$ ways to arrange them. Since this is true for 99 slots out of 100, you can use the well-known summation formula: $$98! \cdot 99 + 98! \cdot 98 + \ldots 98! \cdot 1 = 98!(1+2+ \ldots 99) = 98! \cdot \frac{99 \cdot 100}{2} = 98! \cdot 4950$$