permutations counting techniques, part 1. permutations recall the notion of calculating the number...

Permutations

Counting Techniques, Part 1

Permutations

Recall the notion of calculating the number of ways in which to arrange

a given number of items.

One tool for accomplishing this task is the Multiplication Principle.

Permutations

One way to state the multiplication principle is that to determine the

possible number of outcomes in a given situation, multiply together the number

of options available every time you make a choice.

For example, say you want to order a sundae with 10 flavors of ice cream and 3 different toppings available.

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For example, say you want to order a sundae with 10 flavors of ice cream and 3 different toppings available.

The total possible number of sundaes using one scoop of ice cream and one

topping is 10*3, or 30.

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Another way to state the multiplication principle is this:

If event A can take place in m ways, and event B can take place in n ways, then event A and event B can take place in

m*n ways.

Example: Consider a game where the player flips a coin and rolls a die.

PermutationsExample: Consider a game where the

player flips a coin and rolls a die.How many outcomes are possible for

flipping a coin and rolling a die?

There are 2 possible outcomes for the coin (Heads or Tails) and 6 possible

outcomes for the die (1-6), so there are 2*6, or 12 possible outcomes for the

two events considered together.

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In some instances, determining the possible number of outcomes

requires the use of the Addition Principle.

Here’s one statement of the addition principle:

If event A can occur in m ways and event B can occur in n ways, then event

A or event B can occur in m + n ways.

Permutations

Comparing the multiplication and addition principles:

A school club consists of 14 Juniors and 18 Seniors. In how many ways could

we select a committee of 2 students if one must be a Junior and one must be a

Senior?We have 14 options for the Junior, and 18 options for the Senior, so we have a

total of 14*18, or 252 possibilities.

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A school club consists of 14 Juniors and 18 Seniors. How many options do we

have for a representative if either a Junior or Senior can fill the position?

We have 14 options for a Junior, and 18 options for a Senior, so we have a total

of 14 + 18, or 32 possibilities.

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Recall that in some cases we needed to modify the multiplication principle

—namely when order does not matter. For example, when picking two numbers from a list of 10, we have (10*9)/2, or 45 possibilities.

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We needed to divide by 2 to take into account the fact that we had

counted every pair of numbers twice.I.e., we counted (1, 2) as well as (2, 1) even though they are the same pair of numbers, and should only count

once.

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Likewise, we sometimes need to modify the addition principle.

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Consider a committee that requires either a soccer player or a track athlete. For this example, say the athletes are in a school

where they cannot compete in both soccer and track, due to the two sports being in the

same season.

If there are 40 soccer players and 60 track athletes, how many possibilities are there to select a representative?

Permutations

Because an athlete cannot compete in both sports in this case:

We have 40 + 60, or 100 possibilities.

Soccer Track

40 60

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Now consider a committee that requires either a soccer player or a

band member.

If there are 40 soccer players and 80 band members, how many

possibilities are there to select a representative?

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In this case, a student could be a member of both groups:

Because of the overlap between the two groups, we cannot simply add 40 + 80 and be done with it. We will need to subtract

out the number of students that are in both.

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In this case, a student could be a member of both groups:

If there are 10 students involved in both groups, then we have 40 + 80 – 10, or 110 possibilities for a representative.

10

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In the case of soccer and track, the two groups are mutually exclusive, or disjoint, because it is not possible to

be part of both groups.

Soccer Track

40 60

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In the case of soccer and band, the two groups are not mutually

exclusive.

We must make our adjustment to the addition principle when events are not

mutually exclusive.

10

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In some instances we can use the multiplication principle and the

addition principle together.

Say there is a lottery where a player can arrange (order matters) either 2, 3, or 4

numbers from the numbers 1-20. (Naturally the player who matches 4 numbers wins more money than the

player who only matches 2 numbers.)

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In how many ways can a player fill out a card?

First, consider the number of ways to arrange 2 numbers from the numbers 1-20:

We have 20 choices for the first number, and then 19 for the second number, so we have

20*19, or 380 possibilities.

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The number of ways to arrange 3 numbers:

We have 20 choices for the first number, 19 for the second number, and

then 18 for the third number, so we have 20*19*18, or 6,840 possibilities.

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The number of ways to arrange 4 numbers:

20*19*18*17 = 116,280 possibilities.

PermutationsIn how many ways can a

player fill out a card?

Finally, to determine the number of ways to fill out a card, we need to add together

the possibilities for 2 numbers, for 3 numbers, and for 4 numbers:

380 + 6,840 + 116,280 = 123,500 ways.Can you see why the person matching 4

numbers would win so much more money than someone matching 2 numbers?

Permutations

We have now seen several examples of what are called permutations.

Permutations are the arrangements of a group of items where order matters. We often calculate permutations in a scenario by using the multiplication

principle.

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Example: In how many ways can choose to watch 3 movies?

You have 3 options for your first movie. After choosing the first movie, you will

have 2 options remaining for the second movie. After choosing the first and second movies, only 1 option remains for the third

movie.This gives you 3*2*1, or 6 ways to arrange

3 movies.

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Example: In how many ways can a baseball manager arrange a hitting

order for 9 batters?

The manager has 9 options for the leadoff hitter, then 8 options for the second hitter, then 7 options for the

third hitter, and so forth. In total, the manager has

9*8*7*6*5*4*3*2*1, or 362,880 possible lineups.

PermutationsNotice that the answers can get quite

large, very quickly.Example: How many seating charts

could a teacher make for 20 students, if there are 20 desks in the room?

The teacher has 20 options for the first desk, then 19 options for the second desk, then 18 options for the third desk, and so

forth. In total, the teacher has20*19*18*17*…*5*4*3*2*1,

or 2,432,902,008,176,640,000 possibilities.

PermutationsNotice that the answers can get quite

large, very quickly.Example: How many seating charts

could a teacher make for 20 students, if there are 20 desks in the room?

We usually represent a number this large by using scientific notation:

2,432,902,008,176,640,000might be rounded to 2.43*1018.

Permutations

When dealing with extended multiplications as in the previous

examples, wouldn’t it be nice to have a shortcut?

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When dealing with extended multiplications as in the previous

examples, wouldn’t it be nice to have a shortcut?

Great news—we do!

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Do you remember using factorials back in your days of studying

Algebra? We can use them again in these problems.

Example: The number of ways to arrange 3 movies:

3*2*1 = 3! (read 3 factorial)

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Example: Arranging a batting order for 9 players:

9*8*7*6*5*4*3*2*1 = 9!

Example: Seating charts for 20 students with 20 desks:

20*19*18*17*…*5*4*3*2*1 = 20!

PermutationsOne reason this notation is so useful is

because most calculators (as well as computer spreadsheets) have a

factorial function included, making the calculations quite simple.

PermutationsOn a typical graphing calculator, look

for a button labeled “MATH.”After pressing that button, look for a

heading labeled “PROBABILITY” or something like “PRB” for short. You might need to use the right arrow to

get to that section.Once in the probability section, look

for the factorial function (!).

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To determine the value of 9! using the calculator follow these steps:

•Enter 9•Press MATH•Move to the probability section•Move the cursor to !•Press ENTER—you should see 9! •Press ENTER again to get the answer.

Permutations

The previous examples of permutations (movies, batting orders, and seating charts) share a common feature that makes them relatively

simple: they arranged all of the available items.

This won’t always be the case.

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Example: You want to pick 4 movies out of 10, and then select an order for

viewing them.

In this case, we cannot simply say the answer is 4!, because we are choosing

from more than 4 movies.

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viewing them.

We have 10 options for the first movie. After picking that, we have 9 options for

the second movie, 8 options for the third movie, and finally 7 options for the

fourth movie.

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viewing them.

This gives us10*9*8*7, or 5,040 total possibilities.

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viewing them.

When only ordering 4 items, this is no headache. With a larger number of

items, though, it would be nice to have a more compact formula.

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viewing them.

Notice that we can rewrite 10*9*8*7 as

10*9*8*7*6*5*4*3*2*16*5*4*3*2*1

(We multiply by 6*5*4*3*2*1, and then divide by the same amount.)

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viewing them.

This sneaky little trick allows us to rewrite 10*9*8*7 as

10!6!

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viewing them.

Notice the numerator is the factorial of the number of items available, while the

denominator is the factorial of the difference between the number of items available and the

number of items we’re arranging.

10!6!

__10!__ (10-4)!

=

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In general, if we have n items and want to arrange m of them, we will

have __n!__(n-m)!

permutations.

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Several other notations are commonly used to express this value:

P(n, m)

nPm

PermutationsGraphing calculators often use a function

labeled nPr

The user enters the total number of items, finds the nPr function, and then

enters the number of items to be arranged.

(Example to follow.)

Permutations

Example: A baseball manager wants to arrange a batting lineup for 9 players, and has 15 players from

which to choose. How many batting orders can the manager create?

One solution would be to multiply the individual options:

15*14*13*12*11*10*9*8*7

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Another solution, and one which will prove useful in problems with larger numbers,

would be to use our formula for permutations:

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P(15, 9) =

__15!__(15-9)!

15! 6!

=

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Using the permutation function on a graphing calculator:

•Enter 15 (the total number of players)•Press MATH•Move the cursor to the PRB section•Move the cursor down to nPr •Press ENTER•Enter 9 (the number of players to arrange)•Press ENTER to get the answer

Permutations

Do you remember how the number of seating charts for 20 students was so large?

Now consider a case where there are still 20 students, but there are 25 desks in the room. How many seating charts are now

possible?

Go ahead and guess!

Permutations

Do you remember how the number of seating charts for 20 students was so large?

Now consider a case where there are still 20 students, but there are 25 desks in the room. How many seating charts are now

possible?

P(25, 20) = 1.29*1023 or 129,000,000,000,000,000,000,000

permutations counting techniques, part 1. permutations recall the notion of calculating the number...

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