Question 1131375
there are 2 choices for the first position.
there are 1 choice for the second position.
there are 3 choices for the third position.
there are 2 choices for the fourth position.
there are 1 choice for the fifth position.


the total possibilities are 2 * 1 * 3 * 2 * 1 = 12


those possibilities are:


NR125
NR152
NR215
NR251
NR512
NR521


RN125
RN152
RN215
RN251
RN512
RN521


breaking it down, there are 2 letters that can be in the first two positions in any order, and there are 3 numbers that can be in the last 3 positions in any order.


that gets you 2! * 3! = 2 * 1 * 3 * 2 * 1 = 12 possible arrangements.