# Thread: Combinatorics Problem; Due Friday...

1. ## Combinatorics Problem; Due Friday...

I just can't seem to get this problem, and time is running out.

The problems on this page all deal with hands taken from a deck of 32 cards:
the 7, 8, ... , Q, K, A of each of the four suits.

How many five-card hands are possible which satisfy:

the hand has exactly two hearts
the hand has exactly two spades
the hand has exactly two Kings

For this problem I set up the three cases like this: (O = non-heart, non-spade suits)

1KH - 7H - 1KS - 7S - 14O
1KH - 7H - 7S - 6S - 1KO
7H - 6H - 1KS - 7S - 1KO
Then I multiplied and added them together and got the

However, I can't seem to figure out the second and third
problems.

How many 6-card hands are possible which satisfy:
the hand has exactly two hearts
the hand has exactly three spades
the hand has exactly 2 Kings

I tried setting up the cases the same as in problem #1, with
the necessary modifications:

1KH - 7H - 1KS - 7S - 6S - 14O
1KH - 7H - 7S - 6S - 5S - 1KO
7H - 6H - 1KS - 7S - 6S - 1KO

But My answer of 7350 is wrong and I can't figure out what
I'm doing wrong.

Similarly, I can't figure out the 3rd problem:

How many 6-card hands are possible which satisfy:
the hand has exactly two hearts
the hand has exactly three spades

I would think it would simply be:

8H - 7H - 8S - 7S - 6S - 16O multiplied together, but it isn't.

Any Help would be great.

2. Hello, shaehl!

Surely, you're familiar with Combinations and Factorials ?!

The problems on this page all deal with hands taken from a deck of 32 cards:
the 7, 8, ... , Q, K, A of each of the four suits.

1) How many five-card hands are possible which satisfy:
. . (a) the hand has exactly two Hearts
. . (b) the hand has exactly two Spades
. . (c) the hand has exactly two Kings
(a) Two Hearts

There are 8 Hearts and 24 Others.
We want 2 Hearts and 3 Others.

To get 2 Hearts, there are: .$\displaystyle {8\choose2} = 28$ ways.
To get 3 Others, there are: .$\displaystyle {24\choose3} = 2,024$ ways.

Therefore: .$\displaystyle n(2 \heartsuit) \:=\:28 \times 2,024 \:=\:\boxed{56,672}$

The reasoning (and the answer) is identical to part (a).

(c) Two Kings

There are 4 Kings and 28 Others.
We want 2 Kings and 3 Others.

To get 2 Kings, there are: .$\displaystyle {4\choose2} \:=\:6$ ways.
To get 3 Others, there are: .$\displaystyle {28\choose3} \:=\:3,276$ ways.

Therefore: .$\displaystyle n(\text{2 Kings}) \:=\:6 \times 3,276 \:=\:\boxed{19,656}$

2) How many 6-card hands are possible which satisfy:
. . (a) the hand has exactly two Hearts
. . (b) the hand has exactly three Spades
. . (c) the hand has exactly 2 Kings
(a) Two Hearts

There are 8 Hearts and 24 Others.
We want 2 Hearts and 4 Others.

To get 2 Hearts, there are: .$\displaystyle {8\choose2} \:=\:28$ ways.
To get 4 Others, there are: .$\displaystyle {24\choose4} \:=\:10,626$ ways.

Therefore: .$\displaystyle n(2 \heartsuit) \:=\:28 \times 10,626 \:=\:\boxed{297,528}$

There are 8 Spades and 24 Others.
We want 3 Spades and 3 Others.

To get 3 Spades, there are: .$\displaystyle {8\choose3} \:=\:56$ ways.
To get 3 Others, there are: .$\displaystyle {24\choose3} \:=\:2,024$ ways.

Therefore: .$\displaystyle n(2 \spadesuit) \:=\:56 \times 2,024 \:=\:\boxed{113,344}$

(c) Two Kings

There are 4 Kings and 28 Others.
We want 2 Kings and 4 Others.

To get 3 Kings, there are: .$\displaystyle {4\choose2} \:=\:6$ ways.
To get 4 Others, there are: .$\displaystyle {28\choose4} \:=\:20.475$ ways.

Therefore: .$\displaystyle n(\text{2 Kings}) \:=\:6 \times 20,475 \:=\:\boxed{122,850}$

3) How many 6-card hands are possible which satisfy:
. . (a) the hand has exactly two Hearts
. . (b) the hand has exactly three Spades
Is there a typo?
This is identical to problem #2.

3. Thank you for the thorough response, however, the problems are meant to satisfy all three of the conditions at once. For number 3, we exclude the 2 kings requirement.