Question 253885
Since order matters, he has 5 choices for the first patient, 4 for the second (since he can't see the same person twice), 3 for the third, 2 for the fourth, and one for the fifth.


Multiply these choices out to get 5*4*3*2*1=120



So there are 120 different possibilities. 



Alternatively, you can use the permutation formula {{{P(n,r)=n!/(n-r)!}}} with n=5 and r=5 to get {{{P(5,5)=5!/(5-5)!=5!/0!=(5*4*3*2*1)/1=120/1=120}}} and you'll get the same answer