Statistics Tutorial: Counting Data Points
The solution to many statistical experiments involves being able to count the
number of points in a sample space. Counting points can be hard, tedious, or
both.
Fortunately, there are ways to make the counting task easier. This lesson
focuses on three rules of counting that can save both time and effort - event
multiples, permutations, and combinations.
Event Counter
Use Stat Trek's Event Counter to compute factorials and to count
event multiples, permutations, and combinations. The Event Counter is
free and easy to use. It can be found under the Stat Tools
tab, which appears in the header of every Stat Trek web page.
Event Multiples
The first rule of counting deals with event multiples. An event multiple
occurs when two or more independent events are grouped together. The
first rule of counting helps us determine how many ways an event multiple can
occur.
Rule 1. Suppose we have k independent
events. Event 1 can be performed in n1 ways; Event 2, in n2
ways; and so on up to Event k (which can be performed in nk ways).
The number of ways that these events can be performed together is equal to n1n2
. . . nk ways.
Example 1
How many sample points are in the sample space when a coin is flipped 4 times?
Solution: Each coin flip can have one of two outcomes - heads or tails.
Therefore, the four coin flips can land in (2)(2)(2)(2) = 16 ways.
Example 2
A business man has 4 dress shirts and 7 ties. How many different shirt/tie
outfits can he create?
Solution: For each outfit, he can choose one of four shirts and one of
seven ties. Therefore, the business man can create (4)(7) = 28 different
shirt/tie outfits.
Permutations
Often, we want to count all of the possible ways that a single set of objects
can be arranged. For example, consider the letters X, Y, and Z. These letters
can be arranged a number of different ways (XYZ, XZY, YXZ, etc.) Each of these
arrangements is a permutation.
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In general, n objects can be arranged in n(n - 1)(n
- 2) ... (3)(2)(1) ways. This product is represented by the symbol n!,
which is called n factorial. (By convention, 0! = 1.)
-
A permutation is an arrangement of all or part of a set of objects, with
regard to the order of the arrangement. This means that XYZ is considered
a different permutation than ZYX.
-
The number of permutations of n objects taken r at a time is
denoted by nPr.
Rule 2. The number of permutations of n
objects taken r at a time is
nPr = n(n - 1)(n
- 2) ... (n - r + 1) = n! / (n - r)!
Example 1
How many different ways can you arrange the letters X, Y, and Z? (Hint:
In this problem, order is important; i.e., XYZ is considered a
different arrangement than YZX.)
Solution: One way to solve this problem is to list all of the possible
permutations of X, Y, and Z. They are: XYZ, XZY, YXZ, YZX, ZXY, and ZYX. Thus,
there are 6 possible permutations.
Another approach is to use Rule 2. Rule 2 tells us that the number of
permutations is n! / (n - r)!. We have 3 distinct objects so n = 3. And we want
to arrange them in groups of 3, so r = 3. Thus, the number of permutations is
3! / (3 - 3)! or 3! / 0!. This is equal to (3)(2)(1)/1 = 6.
Example 2
In horse racing, a trifecta is a type of bet. To win a trifecta bet, you need
to specify the horses that finish in the top three spots in the exact order in
which they finish. If eight horses enter the race, how many different ways can
they finish in the top three spots?
Solution: Rule 2 tells us that the number of permutations is n! / (n - r)!. We have 8
horses in the race. so n = 8. And we want to arrange them in groups of 3, so r
= 3. Thus, the number of permutations is 8! / (8 - 3)! or 8! / 5!. This is
equal to (8)(7)(6) = 336 distinct trifecta outcomes. With 336 possible
permutations, the trifecta is a difficult bet to win.
Combinations
Sometimes, we want to count all of the possible ways that a single set of
objects can be selected - without regard to the order in which they are
selected.
-
A combination is a selection of all or part of a set of objects, without
regard to the order in which they were selected. This means that XYZ is
considered the same combination as ZYX.
-
The number of combinations of n objects taken r at a time is
denoted by nCr.
Rule 3. The number of Combinations of n
objects taken r at a time is
nCr = n(n - 1)(n
- 2) ... (n - r + 1)/r! = n! / r!(n - r)! = nPr / r!
Example 1
How many different ways can you select 2 letters from the set of letters: X, Y,
and Z? (Hint:
In this problem, order is NOT important; i.e., XY is considered the same
selection as YX.)
Solution: One way to solve this problem is to list all of the possible
selections of 2 letters from the set of X, Y, and Z. They are: XY, XZ, and YZ.
Thus, there are 3 possible combinations.
Another approach is to use Rule 3. Rule 3 tells us that the number of
combinations is n! / r!(n - r)!. We have 3 distinct objects so n = 3. And we
want to arrange them in groups of 2, so r = 2. Thus, the number of combinations
is 3! / 2!(3 - 2)! or 3! /2!1!. This is equal to (3)(2)(1)/(2)(1)(1) = 3.
Example 2
Five-card stud is a poker game, in which a player is dealt 5 cards from an
ordinary deck of 52 playing cards. How many distinct poker hands could be
dealt? (Hint:
In this problem, the order in which cards are dealt is NOT important;
For example, if you are dealt the
ace, king, queen, jack, ten of spades, that is the same as being dealt
the ten, jack, queen, king, ace of spades.)
Solution: For this problem, it would be impractical to list all of the
possible poker hands. However, the number of possible poker hands can be easily
calculated using Rule 3.
Rule 3 tells us that the number of combinations is n! / r!(n - r)!. We have 52
cards in the deck so n = 52. And we want to arrange them in groups of 5, so r =
5. Thus, the number of permutations is 52! / 5!(52 - 5)! or 52! / 5!47!. This is
equal to 2,598,960 distinct poker hands.
Hint: Although this problem is not hard to solve,
it does involve a lot of computations. You can use Stat Trek's
Event Counter
to do the actual computations. The tool is free, and it reduces
the drudgery.
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