2023-09-27
Define permutations and combinations
Characterize difference between sampling with and without replacement
Characterize difference between sampling when order matters and when order does not matter
Calculate the probability of sampling any combination of the following: with or without replacement and order does or does not matter
Example 1
Suppose we have 10 (distinguishable) subjects for study.
How many possible ways are there to order them?
How many ways to order them if we can reuse the same subject and
need 10 total?
need 6 total?
How many ways to order them without replacement and only need 6?
How many ways to choose 6 subjects without replacement if the order doesn’t matter?
Suppose we have 10 (distinguishable) subjects for study.
Example 1.1
How many possible ways are there to order them?
Example 1.2
How many ways to order them if we can reuse the same subject and
need 10 total?
need 6 total?
Suppose we have 10 (distinguishable) subjects for study.
Example 1.3
How many ways to order them without replacement and only need 6?
Example 1.4
How many ways to choose 6 subjects without replacement if the order doesn’t matter?
Definition: Permutations
Permutations are the number of ways to arrange in order \(r\) distinct objects when there are \(n\) total.
\[nPr = \frac{n!}{(n-r)!}\]
Definition: Combinations
Combinations are the number of ways to choose (order doesn’t matter) \(r\) objects from \(n\) without replacement.
\[nCr = \textrm{"n choose r"} = \binom{n}{r} = \frac{n!}{r!(n-r)!}\]
\[\binom{n}{r} = \binom{n}{n-r}\]
\[\binom{n}{1} = n\]
\[\binom{n}{0} = 1\]
Example 2
Suppose we draw 2 cards from a standard deck without replacement. What is the probability that both are spades when
See table on pg. 277 of textbook
\(n\) = total number of objects
\(r\) = number objects needed
| with replacement | without replacement | |
|---|---|---|
| order matters | \[n^r\] | \[nPr = \frac{n!}{(n-r)!}\] |
| order doesn’t matter | \[ \binom{n+r-1}{r}\] | \[nCr = \binom{n}{r} = \frac{n!}{r!(n-r)!}\] |
Recall, \(P(A) = \dfrac{|A|}{|S|}\). And within combinatorics, we can use the previous equations to help enumerate the event and sample space.
I left something out though… the enumeration of the event is not just one of the above formulas.
For example in the example of the spades when order does not matter, we actually need to enumerate the other cards that were NOT spades. So the event is choosing 2 spades out of 13 AND choosing 0 other cards of 39 cards (13 hearts + 13 clubs + 13 diamonds).
Thus the probability is actually:
\[ P(\text{two spades}) = \dfrac{{13 \choose 2}{39 \choose 0}}{{52 \choose 2}} \]
Chapter 22 Slides