We have the following:
#color(purple)(66*65*64)*color(steelblue)(27*26)#
What we have in purple seems to be #66!#, but it stops at #64#. How do we account for this?
The key realization is that #66!# would be
#66*65*64*color(red)(63*62...)#, but we don't want those extra terms.
If we divide #66!# by what's in red (essentially #63!#), we will be left with #66*65*64#.
#color(purple)((66!)/(63!))=(66xx65xx64xxcancel(63xx62...))/(cancel(63xx62xx61xx60xx59...))#
We can do the same exercise with #27*26#. We just want the first two terms of #27!#, so we divide this #(27-2)!#, or #25!#.
#color(steelblue)((27!)/(25!))=(27xx26xxcancel(25xx24...))/cancel(25xx24xx23xx22...)#
We can rewrite #color(purple)(66*65*64)*color(steelblue)(27*26)# as #color(purple)((66!)/(63!))*color(steelblue)((27!)/(25!))#.
Hope this helps!
