# What is the probability of getting a hand of two clubs and three diamonds?

Nov 20, 2015

(13/52xx12/51xx13/50xx12/49xx11/48)xx""^5C_2 = 267696/31187520

$\approx .008583433373349339$

That is about $1$ in $116$

#### Explanation:

The probability of being dealt two clubs then three diamonds is:

$\frac{13}{52} \times \frac{12}{51} \times \frac{13}{50} \times \frac{12}{49} \times \frac{11}{48}$

But we don't mind what order we get these cards, so this probability needs to be multiplied by ""^5C_2 = (5!)/(2!3!) = (5xx4)/2 = 10 to represent the number of possible orders of clubs and diamonds.