Suppose I have 5 cards, 3 of them are kings and other are 2 queens. I lay them down side by side.
What's the probability of those 3 king cards not being together?
Here's what I did (supposedly incorrect):
- 5! = 120 possible combinations
Pking * Psuits * 3! <-- (Possible positions occupied by the kings)
= 3! * 2! * 3!
= 72 favorable cases
72 / 120 = 6/10 = 0.6
According to my book, this answer seems to be incorrect. Where have I gone wrong? Thanks.
If "not being together" also means that two of the kings can't be together, then the answer should be $$\frac{3!\cdot 2!}{5!} = \frac{1}{10}$$ because when there is no king together with another, there is only one placement of the cards, which is $KSKSK$ where $K$ stands for king and $S$ stands for suit. Then $K$'s can switch places among themselves with $3!$ and so $S$'s do with $2!$. Then the result follows.