This is an example of a problem that involves Permutation with Repetition.
Notice that we have 7 available parking spaces for only 5 bicycles. This means that if we call our 5 bicycles B1, B2, B3, B4, and B5, the following are just few examples that are distinct from each other:
B1-B2-B3-B4-B5-space-space
B1-space-B2-B3-B4-space-B5
space-B1-B2-B3-B4-B5-space
In other words, aside from the 5 bicycles we are arranging, we are also taking into account the arrangement of the 2 extra spaces.
So really, we are arranging 7 things here with two objects (the spaces) being alike. This can be solved by translating it as
[tex] \frac{7!}{1!1!1!1!1!2!} [/tex] where
7! represents the 7 things we are arranging
the five 1!'s represent each of the distinct 5 bicycles; and
2! represents the 2 spaces that are identical.
Therefore, the number of ways that 5 bicycles can parked in parking lot with 7 spaces is [tex] \frac{7!}{1!1!1!1!1!2!}= \frac{7!}{2!}= \frac{(7)(6)(5)(4)(3)(2)(1)}{(2)(1)}=2520 [/tex] ways.
