排列與組合問題 questions?

5.
(a)
Draw 5 cards without restrictions, the number of combinations = ₅₂C₅

To get no pairs :
Firstly, from the 13 numbers (A to K), choose 5 different numbers (13C5)
Second, for each number, choose 1 suit from the 4 suits (4⁵)

P(no pairs)
= 13C5 × 4⁵ / ₅₂C₅
= (13!/5!8!) × 4⁵ / (52!/5!47!)
= 2112/4165

(b)
P(at least one pair)
= 1 - P(no pair)
= 1 - (2112/4165)
= 2053/4165