Data Interpretation & Probability

Permutations and Combinations

Permutations

Definition of n factorial

The definition of n!, pronounced n factorial is
(n)(n - 1)(n - 2). . . (3)(2)(1)

For example:
3! = (3)(2)(1) = 6
4! = (4)(3)(2)(1) = 24
5! = (5)(4)(3)(2)(1) = 120
6! = (6)(5)(4)(3)(2)(1) = 720

Permutation

A permutation is one of the different arrangements of a group of items where order matters.

When order matters AB BA.

Consider the following:

Given 3 people, Bob, Mike and Sue, how many different ways can these three people be arranged where order matters?

Let BMS stand for the order of Bob on the left, Mike in the middle and Sue on the right.

Since order matters, a different arrangement is BSM. Where Bob is on the left, Sue is in the middle and Mike is on the right.

If we find all possible arrangements of Bob, Mike and Sue where order matters, we have the following:

BMS,    BSM,    MSB,    MBS,    SMB,    SBM

The number of ways to arrange three people three at a time is:
3! = (3)(2)(1) = 6 ways

Consider the following:

Given 4 people, Bob, Mike, Sue and Alice, how many different ways can these three people be arranged where order matters?

If we find all possible arrangements of Bob, Mike, Sue and Alice where order matters, we have the following:

BMSA,    BMAS,    BSMA,    BSAM,    BAMS,    BASM
MBSA,    MBAS,    MABS,    MASB,    MSBA,    MSAB
SBMA,    SBAM,    SMBA,    SMAB,    SABM,    SAMB
ABMS,    ABSM,    AMBS,    AMSB,    ASBM,    ASMB
There are 24 ways to arrange the four people four at a time, or 4!

Summary
If you want to arrange 3 people in groups of 3 at a time, there are 3! ways to accomplish this task.

If you want to arrange 4 people in groups of 4 at a time, there are 4! ways to accomplish this task.

If you want to arrange n objects in groups of n at a time, there are n! ways to accomplish this task.

Property

There are n! ways to arrange n objects in groups of n at a time.

Now Suppose we want to take these four people and arrange them in groups of three at a time where order matters. The following demonstrates all the possible arrangements.

BMS,    BSM,    BAS,    BSA,    BMA,    BAM
MBS,    MSB,    MAS, MSA, MBA,    MAB
SBM,    SMB,    SBA,    SAB,    SMA,    SAM
ABM,    AMB,    ABS, ASB,    AMS,    ASM

There are 24 ways to arrange 4 objects taken 3 at a time.
The answer again is 24, but for a different reason. To find the value 24 mathematically we must use the permutation formula. The permutation formula is when order matters.

Permutation Formula

The number of permutations of n objects taken r at a time is:

Example 1:
Find the number of ways to arrange 4 people in groups of 3 at a time where order matters.

Solution:

There are 24 ways to arrange 4 items taken 3 at a time when order matters.

___________________________

Example 2:
Find the number of ways to arrange 6 items in groups of 4 at a time where order matters.

Solution:

There are 360 ways to arrange 6 items taken 4 at a time when order matters.

Combinations

Combination

A combination is one of the different arrangements of a group of items where order does not matter.

When order does not matter AB = BA.

Consider the following:
Given 3 people, Bob, Mike and Sue, how many different ways can these three people be arranged where order does not matters?
Since order does not matter, any arrangement with Bob, Mike and Sue is considered the same arrangement.

Therefore the only arrangement is BMS.

Now suppose we want to take four people, Bob, Mike, Sue and Alice, and arrange them in groups of three at a time where order does not matter. The following demonstrates all the possible arrangements.
BMS,    MSA,    BMA,    BSA

There are 4 ways to arrange 4 people in groups of 3 at a time.

Combination Formula

The number of combinations of a group of n objects taken r at a time is:

Example 3:
Find the number of ways to take 4 people and place them in groups of 3 at a time where order does not matter.

Solution:

Since order does not matter, use the combination formula.

There are 4 ways to arrange 4 items taken 3 at a time when order does not matter.

Example 4:
Find the number of ways to take 20 objects and arrange them in groups of 5 at a time where order does not matter.

Solution:

There are 15,504 ways to arrange 20 objects taken 5 at a time when order does not matter.

Example 5:
Is the State Lotteries a combination or a permutation?

Solution:
The answer has to do with if order matters. If order matters, then it is a permutation. If order does not matter then it is a combination. Do you think the numbers on a ticket have to be in the same order as the order in which they became the winning numbers?

In other words, lets say the winning numbers rolled out of the machine in the order of: 1,2,3,4,5,6. Do the numbers on your ticket have to be in this same order to win? Or will any order such as 2,3,1,5,6,4 also be a winning ticket?

The answer is, any order of the winning numbers will produce the winning ticket. Thus the lotteries are combinations.

Factorial Tips with/without the Calculator

Example 6:
How many ways are there to select a subcommittee of 7 members from among a committee of 17?

Solution:
Since it does not matter what order the committee members are chosen in, the combination formula is used.
Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation.

C(17,7)= 19,448

Therefore there are 19,448 different ways to form the committee.
_

Example 7:
Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

Solution:
When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations.

There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used.

C(52,5) = 2,598,960

Therefore there are 2,598,960 different ways to create a five-card hand from a deck of 52 cards.

Student #1:

There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contain three women and one man?

55 60 25 192

Answer

Student #2:

There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contain at least three women?

65 67 20 12

Answer

Example 8:
A school has scheduled three volleyball games, two soccer games, and four basketball games. You have a ticket allowing you to attend three of the games. In how many ways can you go to two basketball games and one of the other events?

Solution:
Since order does not matter it is a combination.

The word AND means multiply.
Given 4 basketball, 3 volleyball, 2 soccer.

We want 2 basketball games and 1 other event. There are 5 choices left.

C(n,r)
C(How many do you have, How many do you want)
C(have 4 basketball, want 2 basketball)C(have 5 choices left, want 1)

C(4,2)C(5,1)

(6)(5) = 30

Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.
_
Example 9:
How many ways are there to deal a five-card hand consisting of three eight's and two sevens.

Solution:
If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed. The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards.
We have 4 eights and 4 sevens.

We want 3 eights and 2 sevens.

C(have 4 eights, want 3 eights)C(have 4 sevens, want 2 sevens)

C(4,3)C(4,2) = 24

Therefore there are 24 different ways in which to deal the desired hand.

Student #3:

How many different 5-card hands include 4 aces?

61 28 48 225

Answer

Example 10:
How many different 5-card hands include 4 of a kind and one other card?

Solution:

We have 13 different ways to choose 4 of a kind: 2's, 3's, 4's, … Queens, Kings and Aces.

Once a set of 4 of a kind has been removed from the deck, 48 cards are left.
Remember OR means add.
The possible situations that will satisfy the above requirement are:
4 Aces and one other card    C(4,4)C(48,1) = 48

or 4 Kings and one other card    C(4,4)C(48,1) = 48

or 4 Queens and one other card    C(4,4)C(48,1) = 48
.
.
.

or 4 twos and one other card    C(4,4)C(48,1) = 48

Total of 624 ways

This problem could also have been calculated as follows.

13C(4,4)C(48,1) = 624

Therefore there are 624 different 5-card hands that have four of a kind, and one other card.

Student #4:

In a local election, there are seven people running for three positions. The person that has the most votes will be elected to the highest paying position. The person with the second most votes will be elected to the second highest paying position, and likewise for the third place winner. How many different outcomes can this election have?

301 258 176 210

Answer

Student #5:

How many different ways can a five-question true-false test can be answered?

32 28 76 26

Answer

Example 11:
An exacta in horse racing is when you correctly guess which horses will finish first and second. If there are eight horses in the race, how many different possible outcomes for the exacta are there?

Solution:
Does order matter? Yes it does! If the order did not matter who would come to the event? Since order matters, use the formula for a Permutation.
There are 8 horses, and we want to know how many different ways there are to choose 2 horses at a time where order matters.    In other words, once you choose two horses, this counts as two arrangements. If horse A comes in first and horse B comes in second, it is a different arrangement then if horse B comes in first and horse A comes in second.

P(8,2) = 56

Therefore there are 56 different ways to have a first and second place winner out of eight horses.

_
Example 11:
Find the number of ways to draw a straight, (suit does not matter) beginning with a 4 and ending with a 8.

Solution:
There are 5 slots.

The first slot must be a four. There are 4 ways to put a four in the first slot. There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.
(4)(4)(4)(4)(4) = 4⁵ = 1024

Therefore there are 1024 different ways to produce the desired hand of cards.

Student #6:

How many ways can 10 people be placed in alphabetical order according to their first names?

1 10 100 10!

Answer

Student #7:

How many different ways can the letters in the word "store" be arranged?

15 5 120 24

Answer

Example 12:

A certain marathon had 50 people running for first prize, second, and third prize.
A) How many different possible outcomes are there for the first three runners to cross the finish line?

Solution:
Order matters in a race, so use the permutation formula.

P(50,3) = 117,600    ways

B) How many ways are there to correctly guess the first, second, and third place winners?

Solution:
There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.

_

Example 13:

A local delivery company has three packages to deliver to three different homes.
If the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?

Solution:
The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.

We can calculate each of these cases and then add them together, or approach this problem from a different angle.

The only case which is left out of the condition is the case where no wrong packages are delivered.

If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainer will be the three cases above.

There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package. The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.

111 = 1

Determine the total number of ways the three packages can be delivered.

321 = 6

The number of ways at least one house gets the wrong package is:

6 - 1 = 5

Therefore there are 5 ways for at least one house to get the wrong package.

_

Generalized Perms & Combs

In the previous section we dealt with selections without repetitions.    In this section we look at problems that deal with repetitions.

Consider the word fort. Since there are 4 letters grouped 4 at a time where order matters, there are 4! ways to arrange the letters in the word fort.

Note: Order matter when letters are rearranged. In other words, the arrangement of the word fort is a different word when it is arranged as trof.

But what about a word like MISSISSIPPI? How many different arrangements are possible with duplicate letters? The letter S is still S no matter which one you use, so all the words that have duplicate letters such as S, I and P switched are the same words to us.
The word Mississippi is the same word to us a Mississippi. The two letters switched places, but it does not produce a new word for us. Thus, all of these cases can NOT be counted as an arrangement.

Theorem:

Suppose that a sequence S of n items has n1 identical objects of type 1, n2 identical objects of type 2, ni identical objects of type i.    Then the number of orderings of S is:

______________________

The word MISSISSIPPI has the following number of orderings:

= 34,650

The first 4! in the denominator is due to the letter I repeated 4 times.
The second 4! in the denominator is due to the letter S repeated 4 times.
The 2! in the denominator is due to the letter P repeated 2 times.

Student #8:

How many different ways are there to arrange the word school?

455 53 487 360

Answer

Homework Problems

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Use the subject D2.2

Student #1:    The answer is B:

Solution:
Since no order to the committee is mentioned, a combination instead of a permutation is used.

Lets sort out what we have and what we want.

Have: 5 women, 6 men.

Want: 3 women AND 1 man.

The word AND means multiply.

    Woman    and     Men
C(have, want)C(have, want)

C(have 5 women, want 3 women)C(have 6 men, want 1 man)

C(5,3)C(6,1) = 60

There are 60 different ways to form the committee under the above conditions.

Go Back to Lesson

Student #2:    The answer is A:

Solution:
Have: 5 women, 6 men.

Want: 3 women AND 1 man OR 4 women and 0 men.

C(have, want)C(have, want) + C(have ,want)C(have, want)

C(have 5 women, want 3 women)C(have 6 men, want 1 man) + C(have 5 women ,want 4 women)C(have 6 men, want 0 men)

C(5,3)C(6,1) + C(5,4)C(6,0)

60 + 5 = 65

Therefore there are 65 different ways to form the committee under the above conditions.

Go Back to Lesson

Student #3:    The answer is C:

Have: 4 aces and 48 other cards.

Want: 4 aces and 1 other card.

C(have 4 aces, want 4 aces)C(have 48 other cards, want 1 other card)

C(4,4)C(48,1)=48

Therefore there are 48 different arrangements.

Go Back to Lesson

Student #4: The answer is D:

This committee has an order! The person with the highest paying position has a higher order over the other two members of the committee, so order matters and a permutation is used.

There are 7 people and we want groups of 3 at a time where order matters.

P(7,3) = 210

Therefore there are 210 different possible ways this committee can be formed.

This is the same as a slot problem without repetition. Since there are three seats to be assigned, there are three slots. The first slot has 7 different ways to assign the seat, the second slot has 6 different ways to assign the seat, and the third slot has 5 different ways to assign the seat.

Therefore there are 210 different ways to form the committee.

Go Back to Lesson

Student #5:    The answer is A:

Solution:
There are five questions, so there are five slots.

Each question can be answered either true of false.    There are 2 ways for each question to be answered.

Therefore there are 32 different ways the test can be answered.

Go Back to Lesson

Student #6:    The answer is A:

Solution:
There is just one way to place 10 people in alphabetical order. You must find the correct alphabetical order and arrange them accordingly.    Therefore 1 way!

Go Back to Lesson

Student #7:    The answer is C:

Solution:
Order to letters matters. If order did not matter then the word cat would have the same meaning to us as the word tac. There are 5 letters and we want to arrange them 5 at a time.

P(5,5) = 120 ways

You can also figure out this problem using a slot problem:
In the first slot there are 5 ways to place a letter. The second slot now has 4 choices, the third slot has 3 choices. etc.

54*321 = 5! = 120

Therefore there are 120 different arrangements of the word store.

Go Back to Lesson

Student #8:    The answer is D:

Solution:

Therefore there are 360 different arrangements of the word school.

Go Back to Lesson