# additive counting principle

• Dec 2nd 2009, 01:12 PM
sophiaroth
our teacher assigned some pre-req questions but i have no idea how to do this one:

how many 13-card bridge hands include either seven hearts or eight diamonds?

i was thinking you would do 52C13 to figure out how many different hands you could have all together.. but then that doesn't really seem to make sense and i wouldn't know where to go from there.

any help would be great, thanks.
• Dec 2nd 2009, 01:22 PM
Plato
Quote:

Originally Posted by sophiaroth
how many 13-card bridge hands include either seven hearts or eight diamonds?

Note that no hand can contain seven hearts and eight diamonds.
So these events are disjoint. Therefore we and just add,

$\binom{13}{7}\cdot\binom{39}{6}+\binom{13}{8}\cdot \binom{39}{5}$
• Dec 2nd 2009, 01:39 PM
sophiaroth
thanks for the reply, but i'm not really sure where you got those numbers from. or the notation you're using with the numbers over top of eachother
• Dec 2nd 2009, 01:50 PM
Plato
Quote:

Originally Posted by sophiaroth
thanks for the reply, but i'm not really sure where you got those numbers from. or the notation you're using with the numbers over top of eachother

$\binom{N}{k}=_N \mathcal{C}_k =\frac{N!}{k!(N-k)!}$
That is standard for combinations. It called binominal coefficient.

Choose seven hearts and six others or choose eight diamonds and five others.