Permutation, Combinatorics Stuck here : there are 100 objects labeled 1, 2,...100. They are arranged in all possible ways. How many arrangements are there in which object 28 comes before object 29.
My approach : Consider object 28 & object 29 , a single object. Now we have a total of 99 objects which can be per mutated in 99! ways . 
But the answer is 4950*98! .
What's wrong with my approach?

You can take object 28 and 29 as alike. So the permutation order i.e. 28 before 29 is not disturbed. This gives no. of arrangements as 100!/2! = 98!*99*100/2=98!*4950
