Question 460530
a)
(13C4*39C1*4)/52C5 four because there are 4 different suits.
=111540/2598960
=.0429
.
b)
4C2*13*12C3*(4)^3/52C5 any one of the 13 ranks and 2 of 4 suits the remaining 3 cards can have any of the 12 remaining ranks and any of the 4 suits.
=1098240/2598960
=.4226
.
Ed