Combinations and
Permutations
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Easy-To-Understand, Graduate-Level Problems --->
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Combinations and Permutations are smaller groupings of objects often
selected from a larger population. Objects in a Combination are selected
simultaneously from the population. Objects in a Permutation are selected
sequentially from the population.
Basic Explanation of
Combinations and Permutations
Difference
Between Combinations and Permutations
Combination Formula
Permutation Formula
Combination Problems
Problem 1: Number of Possible Combinations of Investment Proposals
Problem 2:
Number of Possible Combinations of Newly Opened Offices
Problem 3:
Number of Possible Combinations of Multiple Newly Opened
Offices
Problem 4:
Number of Possible Combinations of Committees
Problem 5:
Number of Possible Combinations of Sub-Groups
Permutation Problems
Problem 6: Number of Possible
Permutations of Delivery Routes
Problem 7:
Number of Possible Permutations of Seating Arrangements
Problem 8:
Number of Possible Permutations of Executive Groups
Problem 9:
Number of Possible Permutations of Book Arrangements
Problem 10:
Number of Possible Permutations of Letter Groups
The concepts of combinations and permutations are
closely related. A typical problem will ask how many
combination or permutation groups containing x number of
objects can be obtained from a larger population
containing n objects.
The major difference between a combination and a
permutation is when the elements of the group are
chosen:
Combinations - Elements are picked simultaneously, all
at once.
Permutations - Elements are picked sequentially, one
after another.
If there is any order to the arrangements, it is a
Permutation
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The Number of Combinations = nPx / x! = n! / [ x! * (n -
x ) ! ]
of n different objects taken x at a time Simultaneously
For example, the number of Combinations of 9 objects
taken 4 at a time Simultaneously =
= nPx / x! = 9P4 / 4! = n! / [ x! * (n - x )! ]
= 9! / [ 4! * (9 - 4 )! ]
= 126
The Number of Permutations = nPx = n! / (n - x ) !
of n different objects taken x at a time sequentially
For example, the number of Permutations of 9 objects
taken 4 at a time Sequentially =
= nPx = 9P4 = n! / (n - x )!
= 9! / (9 - 4 )!
= 3,024
Problem 1: Combinations of
Investment Proposals
Problem: A company is evaluating 6 investment proposals.
If the company selects 3 of the proposals
simultaneously,
how many different groups of three investment proposals
can be selected?
This is a combination problem because the three
investment
proposals are selected simultaneously.
n = 6 = total number of investment proposals available
for
inclusion in each combination group
x = 3 = number of investment proposals that will be
simultaneously
selected to fill each combination group
The number of combinations of n = 6 different investment
proposals selected x = 3 at a time simultaneously equals:
nPx / x!
= 6P3
/ 3!
= n! / [ x! * (n - x )! ]
= 6! / [3! * (6 - 3)! ]
= 20
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. Everything is explained to you in SIMPLE language in the Excel Statistical Master.
Problem 2: Combinations of Newly
Opened Offices
Problem: A consultancy wants to open 4 offices in 10
Northern
states. Each new office will be in a different state.
The offices
will open all at the same time. How many different ways
can
these four offices be situated among the 10 possible Northern
states?
This is a combination problem because the four different
states
are to selected as locations simultaneously.
n = 10 = total number of states available for inclusion
in each
combination group
x = 4 = number of states that will simultaneously be
selected to
fill each combination group
The number of combinations of n = 10 different states
available to selected at x = 4 at a time simultaneously equals:
nPx / x!
= 10P4
/ 4!
= n! / [ x! * (n - x )! ]
= 10! / [ 4! * (10 - 4 )! ]
= 210
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. The Excel Statistical Master is the fastest way for you to climb the business statistics learning curve.
Problem 3: Multiple Combinations
of Newly Opened Offices
Problem: A consultancy wants to open 4 offices in 10
Northern
states, 3 offices in 9 Southern states, and 2 offices in
8 Eastern
states. Each new office will be in a different state and
all
offices will be opened at the same time. How many
different
combinations does the company have to evaluate?
Total number of combinations =
= 10P4
/ 4! * 9P3
/ 3! * 8P2
/ 2!
= 10! / [4! ( 10 - 4 )!] * 9! /
[3! ( 9 - 3 )!] * 8! / [2! (8 -
2)!]
= 210 * 84 * 28
= 493,920
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. The Excel Statistical Master teaches you everything in step-by-step frameworks. You'll never have to memorize any complicated statisical theory.
Problem 4: Combinations of
Committees
Problem: From a group of 10 men and 8 women, a committee
is formed. The committee will have 3 men and 4 women.
How many different ways can this committee of 3 men and
4 women be formed from the overall group of 10 men and
8 women? All committee members are picked at the same
time.
This is combination problem because all committee
members are
picked at the same time. The problem asks how many ways
can
all possible combinations of 3 out of 10 men be combined
with all
possible combinations of 4 out of 8 women?
The Number of Combinations =
Px / x! = n! / [ x! * (n -
x ) ! ]
of n different objects taken x at a time Simultaneously.
Total number of combinations =
= (All possible combinations of men) * (All possible
combinations of women)
= 10P3
/ 3! * 8P4
/ 4!
= 10! / [3! ( 10 - 3 )!] * 8! /
[4! ( 8 - 4 )!]
= 120 * 70
= 8,400
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. The Excel Statistical Master will make you a fully functional statistician at your workplace.
Problem 5: Combinations of
Sub-Groups
Problem: How many ways can a group of 12 people be
divided into one group of 7 and another group of 5?
This is a combination problem because all members of any
one group can be picked simultaneously.
One way to solve the problem would be to determine the
total number of 7-person combinations that can be formed
from 10 people and then multiply that number by the
number
of 5-person combinations that can be formed from the
remaining 5 people.
The Number of Combinations =
nPx / x! = n! / [ x! * (n -
x ) ! ]
of n different objects taken x at a time Simultaneously.
Total number of combinations =
= (total number of combination of 7 out of 12) * (total
combinations of 5 out of remaining 5)
= 12P7
/ 7! * 5P5
/ 5!
= 12! / [7! ( 12 - 7 )!] * 5! /
[5! ( 5 - 5 )!]
= 792 * 1
= 792
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. With the Excel Statistical Master you can do advanced business statistics without having to buy and learn expensive, complicated statistical software packages such as SyStat, MiniTab, SPSS, or SAS.
Problem 6: Permutations of
Delivery Routes
Problem: A milkman makes 7 deliveries on his route. How
many different sequences can he make to complete all 7
stops?
This is a permutation problem because the stops are done
sequentially.
n = 7 = total number of objects initially available for
inclusion in each
permutation group
x = 7 = number of objects that will
sequentially fill
the
permutation group
The number of permutations of n = 7 different stops
taken x = 7 at a time sequentially equals:
nPx
= 7P7
= n! / (n - x )!
= 7! / (7 - 7)!
= 5,040
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. You'll finally have a solid understanding of business statistics with the Excel Statistical Master.
Problem 7: Permutations of
Seating Arrangements
Problem: How many ways can 5 people be seated on a
sofa if only 3 seats are available and the 3 seats are
filled sequentially by the available 5 people?
This is a permutation problem because the elements
of the permutation group are filled
sequentially.
n = 5 = total number of objects initially available for
inclusion in each
permutation group
x = 3 = number of objects that will
sequentially fill
the
permutation group
The number of permutations of n = 5 different people
seated x = 3 at a time sequentially equal:
nPx
= 5P3
= n! / (n - x )!
= 5! / (5 - 3)!
= 60
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. You'll be able to grasp your statistics course a LOT easier with the Excel Statistical Master.
Problem 8: Permutations of
Executive Groups
Problem: A group of 9 people needs to appoint 1 person
to
be group president, another person to be group vice
president, and a third person to be group treasurer.
If the group first votes for the president, then votes
for
the vice president, and finally votes for the treasurer,
how many different executive groups can be created
from the original 9 people?
This is a permutation problem because the elements of
the
permutation group are filled up sequentially.
n = 9 = total number of objects initially available for
inclusion in
each permutation group
x = 3 = number of object that will
sequentially fill the
permutation group
The number of permutations of n = 9 different people
elected x = 3 at a time sequentially equals:
nPx
= 9P3
= 9! / (9 - 3 )!
= 504
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. The Excel Statistical Master is for you if you want to know how to apply statistics to solve real world business problems.
Problem 9: Permutations of Book
Arrangements
Problem: How many ways can 3 books be placed next to
each other on a shelf one at a time?
This is a permutation problem because the elements of
the
permutation group are filled sequentially.
n = 3 = total number of initially objects available for
inclusion
in each permutation group
x = 3 = number of object that will
sequentially fill the
permutation group
The number of permutations of n = 3 different books
placed x = 3 at a time sequentially equals:
nPx
= 3P3
= n! / (n - x )!
= 3! / (3 - 3)!
= 6
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. If you found your statistics book confusing, You'll really like the Excel Statistical Master. Everything is explained in simple, step-by-step frameworks.
Problem 10: Permutations of
Letter Groups
Problem: From the following six letters: A, B, C, D, E,
F, how
many groups of 3 letters can be created if none of the
letters
from the original 6 are repeated in any group?
This is a permutation problem because none of the
letters can be
repeated. When the first letter of one of the
permutation groups
is chosen, there are only five remaining letters to
choose from.
Thus, the elements of the permutation group are filled
sequentially.
n = 6 = total number of objects initially available for
inclusion in each
permutation group
x = 3 = number of objects that will
sequentially fill
the
permutation group
The number of permutations of n = 6 different letters
placed
x = 3 at a time sequentially equally:
nPx
= 6P3
= n! / (n - x )!
= 6! / (6 - 3)!
= 120
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. You'll be able to grasp your statistics course a LOT easier with the Excel Statistical Master.
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