Stat 330 (Spring 2015): slide set 4
P (n, k) =
n!
(n − k)!
P (4, 4) =
4!
(4−4)!
=
4!
0!
=
24
1
= 24.
2
That’s the situation, where we take 4 objects out of a set of 4 and order
them - that is P (4, 4)!.
You only remember that a friend’s (4 digit) telephone number consists of
the numbers 3, 4, 8 and 9. How many different phone numbers are possible?
Example 1:
Theorem:
Permutation Number: P (n, k) - The number of permutations of k objects
taken from n.
Permutation - An ordering of k distinct objects chosen from n distinct
objects.
|Ω| in the previous example has a name . . .
Permutations
Last update: January 13, 2015
Stat 330 (Spring 2015)
Slide set 4
Stat 330 (Spring 2015): slide set 4
n!
(n−k)!
Stat 330 (Spring 2015): slide set 4
1
3
2. One of the choices is “olives”. If all possible rankings are equally likely,
what is the probability that a randomly selected survey has “olives” in
the top 3?
1. How many possible answers to the survey question are there?
A survey question lists seven pizza toppings and asks you to rank your
favorite 3.
Example 2: Pizza Toppings
Permutations: Examples
=
Multiplication principle implies |Ω| = n(n − 1)(n − 2) · · · (n − (k − 1))
n possibilities for x1, n − 1 possibilities for x2, . . ., n − (k − 1) possibilities
for xk .
Break the complex action into a series k single draws.
(i.e.,xi is outcome on draw i).
What is |Ω|?
Ω = {(x1, . . . , xk ) : xi ∈ {1, . . . , n}, xi = xj }
Sample Space:
Experiment: A box has n items numbered 1, . . . , n. Select k(≤ n) items
without replacement. (An item is drawn at most once). Keep track of the
sequence of selections.
Ordered Samples Without Replacement
Stat 330 (Spring 2015): slide set 4
n!
n
C(n, k) =
=
(n − k)!k!
k
6
7
• If the coin is fair, what is the probability of getting 17 heads?
• How many ways are there to get 17 heads?
Toss a coin 23 times.
Example 2: Coin Toss
• What is the probability that you win?
• What is the probability that all 5 match?
The lottery picks 5 numbers from {1, . . . , 49} without replacement. The
order in which the numbers are picked is irrelevant. You win if you pick at
least three of the same numbers that the lottery picks.
Example 1: Lottery (pick-five)
Combinations: Examples
P (n, k)
n!
=
P (k, k) (n − k)!k!
Stat 330 (Spring 2015): slide set 4
|Ω| =
Stat 330 (Spring 2015): slide set 4
Thus
= |Ω|P (k, k).
P (n, k) = (# of ways to select k objects from n)×(# ways to order the k)
Thus,
The left hand side, # of ways to order k objects chosen from n, is P (n, k)
5
Number of Combinations: The number of combinations of k objects chosen
from n is C(n, k)
Theorem:
Stat 330 (Spring 2015): slide set 4
× (# of ways that k distinct objects can be ordered)
(# of ways to select k distinct objects from n objects)
# of ways to order k objects chosen from n =
Hence by multiplication principle,
Continued...
4
Combination - A subset of n distinct objects which has k distinct objects.
|Ω| in the previous example has a name . . .
Combinations
• Order the k objects.
• Select a subset of k objects from n.
It is important to note that the complex action of obtaining an ordering of
k distinct objects from n distinct objects can be broken down to two simple
components:
|Ω| = # of ways to select k distinct objects from n objects.
What is |Ω|?
Ω = {{x1, . . . , xk } : xi ∈ {1, . . . , n}, xi = xj }
Sample Space:
Experiment: A box has n items numbered 1, . . . , n. Select k ≤ n items
without replacement. (A number is drawn at most once). Keep track of
the set of numbers selected. (Order does not matter).
Unordered Samples Without Replacement
C(n, k) =
Unordered Sample Without Replacement
k
10
=
P (n, k) =
Ordered Sample Without Replacement
n!
(n−k)!k!
n!
(n−k)!
nk
Ordered Sample With Replacement
n
Number of Possible Outcomes
Method
Counting Summary
8
13 4
1 · 3
Stat 330 (Spring 2015): slide set 4
The number of ways to do this is (using the multiplication rule) =
• Then select the 3 cards of this rank out of 4 suits.
• Select the rank
Now, break down the selection of the first three cards into the following
two steps:
First, break down to two steps: select the first three cards, then select the
next two cards.
We use the multiplication rule to compute the number of ways a full house
hand may be chosen.
Example:
Stat 330 (Spring 2015): slide set 4
Solution to the “Full House” problem
Thus
P (“Full House”) =
|“Full House”|
=
|Ω|
5
12 4
1 · 2
134 124
1
3 · 1
2
52
≈ .0014
Therefore the number of ways to select a full house hand
4 124
(using the multiplication rule) is: = 13
1
3 · 1
2
(since we have already used one rank for the first 3) =
The number of ways to do this is:
Now, break down the selection of the other two cards likewise.
9
Stat 330 (Spring 2015): slide set 4
Solution to the “Full House” problem (continued)...