# number of pairs

• Jun 18th 2008, 08:17 AM
robkitch
number of pairs
if u are dealt 5 cards what is the possible way of getting a pair which a pair can be a 2 and a 3
• Jun 18th 2008, 08:35 AM
janvdl
Quote:

Originally Posted by robkitch
if u are dealt 5 cards what is the possible way of getting a pair which a pair can be a 2 and a 3

EDIT:
Soroban's post gave me greater insight just now :D

$\displaystyle \frac{ {4 \choose 1} {4 \choose 1}{44 \choose 3} }{{52 \choose 5}} = 0.08153 = 8,15 \ percent \ approx$
• Jun 18th 2008, 08:41 AM
Soroban
Hello, robkitch!

Quote:

if u are dealt 5 cards what is the possible way of getting a pair which a pair can be a 2 and a 3
I'll take a guess at what you meant . . .

We want five cards: one Deuce, one Trey, and three Others.

To get one Deuce, there are: $\displaystyle {4\choose1} = 4$ ways.

To get one Trey, there are: $\displaystyle {4\choose1} = 4$ ways.

To get three Others, there are: $\displaystyle {44\choose3} = 13,244$ ways.

Therefore, there are: .$\displaystyle 4 \times 4 \times 13,244 \:=\:\boxed{211,904\text{ ways}}$