Combinations and Permutations
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  combinations, permutations, and event multiples.
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.

The number of combinations of n objects taken r at a time is
denoted by _{n}C_{r}.
Rule 1. The number of combinations of n
objects taken r at a time is
_{n}C_{r} = n(n  1)(n
 2) ... (n  r + 1)/r! = n! / r!(n  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 1. Rule 1 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}C_{2} = 3! / 2!(3  2)! = 3! /2!1! = (3)(2)(1)/(2)(1)(1) = 3
Example 2
Fivecard 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 1.
Rule 1 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 combinations is:
_{52}C_{5} = 52! / 5!(52  5)! or 52! / 5!47! = 2,598,960
Hence, there are 2,598,960 distinct poker hands.
Combination and Permutation Calculator
Use Stat Trek's Combination and Permutation Calculator to (what else?)
compute combinations and permutations. The calculator is
free and easy to use. You can find the Combination and Permutation Calculator in Stat Trek's
main menu under the Stat Tools tab. Or you can tap the button below.
Combinations and Permutations
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.

The number of permutations of n objects taken r at a time is
denoted by _{n}P_{r}.
Rule 2. The number of permutations of n
objects taken r at a time is
_{n}P_{r} = 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}P_{3} = 3! / (3  3)! = 3! / 0! = (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.
_{8}P_{3} = 8! / (8  3)! or 8! / 5! = (8)(7)(6) = 336
Conclusion: With 336 possible permutations, the trifecta is a difficult bet to win.
How are combinations and permutations related?
Combinations and permutations are related according to the following formulas:
_{n}P_{r} = _{n}C_{r} * r!
and
_{n}C_{r} = _{n}P_{r} / r!
Event Multiples
The third rule of counting deals with event multiples. An event multiple
occurs when two or more independent events are grouped together. The
third rule of counting helps us determine how many ways an event multiple can
occur.
Rule 3. Suppose we have k independent
events. Event 1 can be performed in n_{1} ways; Event 2, in n_{2}
ways; and so on up to Event k (which can be performed in n_{k} ways).
The number of ways that these events can be performed together is equal to n_{1}n_{2}
. . . n_{k} 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.
Event Counter
Use Stat Trek's Event Counter to quickly count event multiples. The Event Counter is
free and easy to use. It can found in the Stat Trek
main menu under the Stat Tools tab. Or you can tap the button below.
Event Counter
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.