Question 934989
5 cards are selected from  10 cards numbered from 1 to 10 w/o replacement
P(1,2,3,_,_) = {{{(1/10)(1/9)(1/8)(7/7)(6/6)}}} = 1/720
Re TY
My thoughts were the Probability depended only on the first three choices...
as the last two had no strings attached to them.
Note: 10P3 = 720... P(1,2,3) = 1/720
P(1,2,3,_,_) = {{{(1/720)((7C2)/(7C2))}}}