# Math Help - Counting (Permutations and Combinations)

1. ## Counting (Permutations and Combinations)

I'm having a little trouble understanding some of this.

This is an example from a practice test

7 – 8. A poker hand consists of 5 cards dealt from a deck of 52 distinct cards.

7. In how many ways a full house (3 of one type of card and 2 of another) be dealt?
a. 168
b. 432
c. 3744
d. 29,952
e. none of these

8. How many different hands contain exactly 3 aces?
a. 1128
b. 4512
c. 1682
d. 18,048
e. none of these

I have .. no idea how to approach this.

There are 13 types of cards. So your chances of drawing a full house almost seems like it would be
4*3*2 * 4*3

But that doesn't make much sense either... gahh!

2. [size=3]Hello, JTG2003!

A poker hand consists of 5 cards dealt from a deck of 52 distinct cards.

7. In how many ways a full house (3 of one type of card and 2 of another) be dealt?

. . $(a)\;168 \qquad(b)\;432 \qquad(c)\;3744\qquad(d)\;29,952\qquad (e)\;\text{none of these}$
We want a Triple and a Pair.

There are 13 choices for the value of the Triple.
There are: . ${4\choose3} = 4$ ways to get the Triple.

There are 12 choices for the value of the Pair.
There are: . ${4\choose2} = 6$ ways to get the Pair.

Therefore, there are: . $13\cdot4\cdot12\cdot6 \:=\:{\color{blue}3744}$ ways to get a Full House.

8. How many different hands contain exactly 3 aces?

. . $(a)\;1128 \qquad (b)\;4512 \qquad (c)\;1682 \qquad (d)\;18,\!048 \qquad (e)\;\text{none of these}$
The deck contains 4 Aces and 48 Others.
. . We want 3 Aces and 2 Others.

There are: . ${4\choose3} = 4$ ways to get 3 Aces.

There are: . ${48\choose2} = 1128$ ways to get 2 Others.

Therefore, there are: . $4\cdot1128 \:=\:{\color{blue}4512}$ ways to get exactly 3 Aces.