Permutation & combination problem,Platform problem, At a platform there are 3 gates numbered 1,2 and 3.In how many ways can 100 people get inside the platform? Given that only 1 may enter through 1 gate at a time. Ans:102!/2! I need the explanation Thanks.

You can order the 100 people in a row, and insert 2 separator between them. Then the people before the first separator will enter through gate 1, the people between the two separators will enter through gate 2, and the other through gate 3.

The number of options to place 2 identical separators and 100 people in a row is $\dfrac{102!}{2!}$.

If the people are identical, the number of options are $\dfrac{102!}{100!2!}={102 \choose2}$