The symbol !! means factorial, which is defined recursively as
n! =n*(n-1)!n!=n⋅(n−1)!
0! =10!=1
Then, 9!9!
=9*8!=9⋅8!
=9*8*7!=9⋅8⋅7!
=9*8*7*6!=9⋅8⋅7⋅6!
=9*8*7*6*5!=9⋅8⋅7⋅6⋅5!
=9*8*7*6*5*4!=9⋅8⋅7⋅6⋅5⋅4!
=9*8*7*6*5*4*3!=9⋅8⋅7⋅6⋅5⋅4⋅3!
=9*8*7*6*5*4*3*2!=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!
=9*8*7*6*5*4*3*2*1*0!=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1⋅0!
=9*8*7*6*5*4*3*2*1*1=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1⋅1
=362880=362880
As seen, the factorial of a certain integer is basically multiplying all integers between 11 and that integer.
""_nP_knPk is defined as (n!)/((n-k)!)n!(n−k)!. This is used to determine the number of ways you can choose kk items from nn items, where the order is important. ""_25P_325P3 is then (25!)/((25-3)!)=(25!)/(22!)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)25⋅24⋅23⋅22⋅21⋅20⋅19⋅18⋅17⋅16⋅15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅122⋅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.
