Counting techniques

When the outcomes of an experiment are equally
likely, computing probability reduces to counting.

If N is the number of outcomes in the sample
space and N(A) the number of outcomes in the
event A, then P(A)=N(A)/N.

If a list of outcomes is easily obtained and N is
small, then N and N(A) may be determined
without the benefit of counting techniques.
However, this isn’t always the case.
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
If an experiment consists of k stages, where
stage j can be carried out in n j ways, the
number of ways to carry out the experiment is
n1n2
nk
3

After selecting an object it is replaced before
the next object is taken

If a sample of size k is taken from a set of n
objects, the number of possible ordered
samples is n k .

Example: If a die is rolled five times, the
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number of ordered samples is .
4

Each of the seven departments has one representative
on the student council. From these seven, one is
selected as chair, another as vice-chair, and a third to
be the secretary. How many ways are there to select
the three officers?

Now suppose three of the seven representatives are to
be selected to attend a convention. In how many ways
can the three be chosen?

In the first situation, order matters, in the second it
doesn’t.
5

Pk ,n
If
denotes the number of ways of
choosing k objects of n, where order matters,
then
n!
Pk ,n  n  n  1 n  2

 n   k  1  
 n  k !
For the selection of chair, vice-chair, and
secretary, there are (7)(6)(5)=7!/(7-3)! =210
permutations.
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
If order doesn’t matter, we have to remember
that each of the Pk ,k  k ! orderings have been
counted separately. We need to divide by this
number, so that
Pk ,n
n
n!
Ck ,n 

k!

 
k ! n  k !  k 
For the selection of the committee to go to
7
the convention, there are  3   35 combinations.
 
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
An iPod playlist contains 100 songs, 10 of which
are by the Beatles. What is the probability that
the first Beatles song heard is the fifth song
played?

The total number of ways to play the first five
songs is 100(99)(98)(97)(96). The number of
these where the first Beatles song is the fifth
song is 90(89)(88)(87)(10).

The answer is the ratio of the two.
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
 100 
 10 


If we don’t consider order, there are
ways to choose the location of the Beatles
songs (the denominator). Of these, we have
 95 
 9  ways to choose the location of the last 9
 
Beatles songs, and one specific way to choose
the first five selections. Taking the ratio, we get
the same answer as before.
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