Assuming the bicycles are distinguishable
there are #6# choices for the first bike;
and for each of these #6# choices there remain #5# choices for the second bike (giving #6xx5=30# choices for the first two bikes);
and for each of these #6xx5# choices there remain #4# choices for the third bike (giving #6xx5xx4=120# choices for the first #3# bikes);
and for each of these #6xx5xx4# choices there remain #3# choices for the fourth bike (giving #6xx5xx4xx3=360# choices for the first #4# bikes);
and for each of these #6xx5xx4xx3# choices there remain #2# choices for the fifth bike (giving #6xx5xx4xx3xx2=720# choices for the first #5# bikes);
and for each of these #6xx5xx4xx3xx2# choices there remains #1# choices (really no choice) for the sixth bike (giving #6xx5xx4xx3xx2xx1=720# choices for the #6# bikes);
Notation: #6xx5xx4xx3xx2xx1# can be written as #6!# (read as "six factorial").