Combinatorics Problem with Bridge

[khalidh223]

A bridge hand (13 cards from a 52-card deck) has high honors if it contains the five highest-ranked cards (10, Jack, Queen, King, Ace) in the same suit. How many bridge hands have high honors?

I know it is a problem involving n choose k, but I am stuck on when to apply them and on what input (or multiple inputs?) of size n.

 

[Dr.Peterson]

Never approach a problem with a tool in your hand.

You might think that combinations are going to be the way to go, but they may not be; you might waste a lot of time trying to figure out how to use your hammer to change that light bulb (and break a lot of bulbs along the way).

Instead, first think about how you might go about listing all hands containing high honors.

You might start by choosing one suit, and taking the five cards from that suit. How many ways are there to do that?

Then you have to fill out the rest of the hand. How many ways can you do that?

Unfortunately, in doing this, you may be counting the same hand more than once, because a hand might have high honors in two or more hands. There are ways to correct for this, but it might be a lot of work. So you may want to look for other ways to count. Possibly there is an easier way to count hands that do not contain high honors, for example.

I haven't even tried to actually solve the problem yet; I want to get you to try things, because that is the only way to really learn combinatorics.

The other reason is that I need to know what you have learned besides combinations (n choose k). Are you just starting this subject, or are there slightly more advanced ideas you could use?

 

[Romsek]

Pick a suit. Select the high honors cards. Four ways to do this.
Now take the remaining 47 cards and select 8. \(\displaystyle \dbinom{47}{8}\) ways to do this.

But those selections of 8 will include some that are also high honors so we are double counting.
We have to subtract off the number of hands that have high honors in 2 suits.

Choose two suits. \(\displaystyle \dbinom{4}{2}\) ways to do this. We can't get further high honors as there
aren't enough cards in the hand so we can choose any 3 cards from the remaining 42 in the deck.
Thus there are \(\displaystyle \dbinom{4}{2}\dbinom{42}{3}\) hands that have high honors in 2 suits.

So we have \(\displaystyle N = 4 \dbinom{47}{8} - \dbinom{4}{2}\dbinom{47}{3} = 1257732690\)