The symbol #!# means factorial, which is defined recursively as
#n! =n*(n-1)!#
#0! =1#
Then, #9!#
#=9*8!#
#=9*8*7!#
#=9*8*7*6!#
#=9*8*7*6*5!#
#=9*8*7*6*5*4!#
#=9*8*7*6*5*4*3!#
#=9*8*7*6*5*4*3*2!#
#=9*8*7*6*5*4*3*2*1!#
#=9*8*7*6*5*4*3*2*1*0!#
#=9*8*7*6*5*4*3*2*1*1#
#=362880#
As seen, the factorial of a certain integer is basically multiplying all integers between #1# and that integer.
#""_nP_k# is defined as #(n!)/((n-k)!)#. This is used to determine the number of ways you can choose #k# items from #n# items, where the order is important. #""_25P_3# is then #(25!)/((25-3)!)=(25!)/(22!)#. Expanding this out using the process similar to the previous problem, we get #(25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)#.
Notice that we can cancel some of these factors: #(25*24*23*cancel(22)*cancel(21)*cancel(20)*cancel(19)*cancel(18)*cancel(17)*cancel(16)*cancel(15)*cancel(14)*cancel(13)*cancel(12)*cancel(11)*cancel(10)*cancel(9)*cancel(8)*cancel(7)*cancel(6)*cancel(5)*cancel(4)*cancel(3)*cancel(2)*cancel(1))/(cancel(22)*cancel(21)*cancel(20)*cancel(19)*cancel(18)*cancel(17)*cancel(16)*cancel(15)*cancel(14)*cancel(13)*cancel(12)*cancel(11)*cancel(10)*cancel(9)*cancel(8)*cancel(7)*cancel(6)*cancel(5)*cancel(4)*cancel(3)*cancel(2)*cancel(1))#. We are left with #25*24*23#, which is equal to #13800#.
