4
votes
1answer
119 views

Combinations which do not have elements in common

I can choose 2 letters from the four letters $\{A,B,C,D\}$ in 6 combinations using the combination formula $$\frac{n!}{ r! (n-r)!}$$ ...
7
votes
3answers
461 views

Probability problem — Rube Goldberg solution?

A user posted this question on StackOverflow which was closed as off topic: 3 people are playing a game with a standard 52 card deck. Each player is given 2 cards each, possible cards and their ...