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In particular, they are called the permutations of five objects taken two at a time, and the number of such permutations possible is denoted by the symbol 5 P 2, read “5 permute 2.” In general, if there are n objects available from which to select, and permutations ( P ) are to be formed using k of the objects at a time, the number of. Permut8 is a plug-in that require a host application that supports VST 2.4 or Audio Unit 2 plug-ins. If you do not have a host application you can still try it out in a free host.

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Permutations and Combinations are super useful in so many applications – from Computer Programming to Probability Theory to Genetics.

I'm going to introduce you to these two concepts side-by-side, so you can see how useful they are.

The key difference between these two concepts is ordering. With Permutations, you focus on lists of elements where their order matters.

For example, I was born in 1977. That's number 1 followed by number 9, followed by number 7, followed by number 7. In that particular order.

If I change the order to 7917 instead, that would be a completely different year. Thus, the order matters.

With Combinations on the other hand, the focus is on groups of elements where the order does not matter.

Like my cup of coffee is a combination of coffee, sugar and water. It doesn't matter which order I add these ingredients are in. There may as well be water, sugar and coffee, it's still the same cup of coffee. Thus, the order does not matter.

Now let's take a closer look at these concepts.

Part 1: Permutations

Permutations Where Repetition is Allowed

Imagine you got a new phone. As you start using this new phone, at some point you will be asked to set up a password.

The password must consist of 4 digits. Any 4 digits. And they may be repeated.

There are 10 digits in total to begin with. Those are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. So for the first digit of your password, you have 10 choices.

Since you may use the same digit again, the number of choices for the second digit of our password will be 10 again! Thus, choosing two of the password digits so far, the permutations are 10 times 10, or 10 x 10 = 100 or 10².

The same thinking goes for the third digit of your password. You get to choose from the same 10 choices again. This time you will have 10 times 10 times 10, or 10 x 10 x 10 = 1,000 or 10³ permutations.

At last, for the fourth digit of the password and the same 10 digits to choose from, we end up with 10 times 10 times 10 times 10, or 10 x 10 x 10 x 10 = 10,000 or 10⁴ permutations.

As you probably noticed, you had 4 choices to make and you multiplied 10 four times (10 x 10 x 10 x 10) to arrive at a total number of permutations (10,000). If you had to choose 3 digits for your password, you would multiply 10 three times. If 7, you would do it seven times, and so on.

But life isn't all about passwords with digits to choose from. What if you have a birthday party and need to choose 5 colored balloons from 20 different colors available?

Since you have 20 different colors to choose from and may choose the same color again, for each balloon you have 20 choices. The first balloon is 20, the second balloon is 20 times 20, or 20 x 20 = 400 etc. For the fifth balloon you get 20 x 20 x 20 x 20 x 20 = 3,200,000 or 20⁵ permutations.

Let's summarize with the general rule: when order matters and repetition is allowed, if n is the number of things to choose from (balloons, digits etc), and you choose r of them (5 balloons for the party, 4 digits for the password, etc.), the number of permutations will equal P = n^r.

Permutations Where Repetition Isn't Allowed

Next, let's consider the case where repetition is not allowed. As an example, we will look at the planets of our solar system.

How many different ways can you arrange these 8 planets? The planets are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune. After choosing, say, Mercury you can't choose it again. Thus, you have to reduce the number of available choices each time the planet is chosen.

The first choice will have 8 possibilities. The second choice will have 8 minus 1 equals 7 possibilities, then 6, followed by 5, followed by 4, until we have 1 planet left in the list.

Following the logic from the previous scenario, the total number of permutations is: P = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320.

In other words, this is a product of integer 8 and all the positive integers below it. This product is called Factorial and is denoted with an exclamation point, like this: 8!

The number of permutations equals P = 8! or more generally P = n!

What if you only need to arrange, say, 5 out of these 8 planets instead of all of them? Then you only take the first 5 steps in our method. Namely, P = 8 x 7 x 6 x 5 x 4 = 6,720 will be how many ways you can arrange 5 planets out of 8.

But why stop here? Why not apply our logic to come up with a more general formula? To make the above notation easy to remember for any numbers of objects, we will use a trick. In a fraction, multiplying both numerator and denominator by the same number (except zero), does not affect that fraction. Thus:

Number of planets to choose from n = 8, you choose r = 5 of them. Substituting the numbers into above formula gives us P = 8! / (8 - 5)! = 8! / 3!. Same as 8 x 7 x 6 x 5 x 4 = 6,720.

From here, the result from earlier example can be derived. There, you arranged all 8 out of 8 available planets. Using the new formula, P = 8! / (8 - 8)! = 8! / 0!. Since, factorial of zero is agreed to equal 1, P = 8! / 1 = 8!. Or more generally:

P = n! / (n - n)! = n! / 0! = n!.

One short and convenient notation often used is: P(n, r) = n! / (n - r)!

Remembering formulas is important. But what's more important for solving real life problems is to know which formulas to use in each situation. Practice helps.

Pop quiz:

The tournament is on and six teams are competing. First place gets gold and second place gets silver medals. How many distinct ways can medals be awarded to these teams?

Choose 1 answer

Explanation: you have 6 teams to choose from. Thus n = 6. Gold and silver together give you 2 medals to award. Thus r = 2. Substituting these numbers into your formula gives us P(6, 2) = 6! / (6 - 2)! = 6! / 4! = 6 x 5 = 30.

Part 2. Combinations

Combinations Without Repetition

To make the comparison more vivid, let's revisit our planet selection example. What if you want to know just which planets are chosen and not their order of appearance?

There you had 6,720 distinct ways of arranging 5 out of 8 planets. But since the order of appearance does not matter now, many of these ways are redundant. They are the same to us.

A group of Venus, Earth, Mars, Jupiter, Saturn is the same group as Mars, Jupiter, Venus, Earth, Saturn and the group as Saturn, Mars, Earth, Jupiter, Venus. These are just different sequences of the same 5 planets.

How many groups do you have that are the same? If you choose r planets per group, you get r! groups. For r = 5, you get r! = 5! = 120 groups.

Thus, to eliminate the unnecessary groups that are the same, you divide the number of original 6,720 Permutations by 5!. The result is 6,720 / 120 = 56.

To generalize, in order to arrive at the number of Combinations, you need to figure out all the Permutations and divide by all the Redundancies.

Using short and convenient notation: C(n, r) = P(n, r) / r! = n! / (r!(n - r)!)

And this assumes that order does not matter and there are no repetitions (that is – there is only one Jupiter to choose from).

Let's revisit the tournament example:

The tournament is on and six teams are competing. First place gets gold and second place gets silver medals. How many groups of medal winners are possible? Order of teams doesn't matter

Choose 1 answer

As before, you have 6 teams. Thus, n = 6. There are two medals awarded, so r = 2. However, this time it doesn't matter who wins gold and who wins silver. Team gold and team silver is the same as team silver and team gold. Substituting these numbers into your formula gives us C(6, 2) = 6! / (2!(6 - 2)!) = 6! / 2! 4! = 15.

Combinations with Repetition

To complete this article, there is one case that requires special attention. So far in our Combinations we assumed there was no repetition. No two items were the same.

What if we can have repetitions? What if, as in our earlier example, we can choose more than one balloon of the same color? If the number of balloons to choose from is n and we choose r of them while allowing for same colors and disregarding the order of arrangement, we will end up with (n + r - 1)! / (r!(n - 1)!) Combinations.

So wrapping up, here is a table you can use to reference these concepts and their formulas.

I hope this article has helped you better understand these two important mathematical concepts. Thanks for reading.

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24

BY THE PERMUTATIONS of the letters abc we mean all of their possible arrangements:

abc

acb

bac

bca

cab

cba

There are 6 permutations of three different things. As the number of things (letters) increases, their permutations grow astronomically. For example, if twelve different things are permuted, then the number of their permutations is 479,001,600.

Now, this enormous number was not found by counting them. It is derived theoretically from the Fundamental Principle of Counting:

If something can be chosen, or can happen, or be done, in m different ways, and, after that has happened, something else can be chosen in n different ways, then the number of ways of choosing both of them is m·n.

For example, imagine putting the letters a, b, c, d into a hat, and then drawing two of them in succession. We can draw the first in 4 different ways: either a or b or c or d. After that has happened, there are 3 ways to choose the second. That is, to each of those 4 ways there correspond 3. Therefore, there are 4· 3 or 12 possible ways to choose two letters from four.

ab means that a was chosen first and b second; ba means that b was chosen first and a second; and so on.

Let us now consider the total number of permutations of all four letters. There are 4 ways to choose the first. 3 ways remain to choose the second, 2 ways to choose the third, and 1 way to choose the last. Therefore the number of permutations of 4 different things is

4· 3· 2· 1 = 24

Thus the number of permutations of 4 different things taken 4 at a time is 4!. (See Topic 19.)

(To say 'taken 4 at a time' is a convention. We mean, '4! is the number of permutations of 4 different things taken from a total of 4 different things.')

In general,

The number of permutations of n different things taken n at a time
is n!.

Example 1. Five different books are on a shelf. In how many different ways could you arrange them?

Answer. 5! = 1· 2· 3· 4· 5 = 120

Example 2. There are 6! permutations of the 6 letters of the word square.

a) In how many of them is r the second letter? _ r _ _ _ _

b) In how many of them are q and e next to each other?

Solution.

a) Let r be the second letter. Then there are 5 ways to fill the first spot. After that has happened, there are 4 ways to fill the third, 3 to fill the fourth, and so on. There are 5! such permutations.

b) Let q and e be next to each other as qe. Then we will be permuting the 5 units qe, s, u a, r.. They have 5! permutations. But q and e could be together as eq. Therefore, the total number of ways they can be next to each other is 2· 5! = 240.

Permutations of less than all

We have seen that the number of ways of choosing 2 letters from 4 is 4· 3 = 12. We call this

'The number of permutations of 4 different things taken 2 at a time.'

We will symbolize this as ⁴P₂:

⁴P₂ = 4· 3

The lower index 2 indicates the number of factors. The upper index 4 indicates the first factor.

For example, ⁸P₃ means 'the number of permutations of 8 different things taken 3 at a time.' And

For, there are 8 ways to choose the first, 7 ways to choose the second, and 6 ways to choose the third.

In general,

ⁿP_k = n(n − 1)(n − 2)··· to k factors

Factorial representation

We saw in the Topic on factorials,

5! is a factor of 8!, and therefore the 5!'s cancel.

Now, 8· 7· 6 is ⁸P₃. We see, then, that ⁸P₃ can be expressed in terms of factorials as

In general, the number of arrangements -- permutations -- of n things taken k at a time, can be represented as follows:

The upper factorial is the upper index of P, while the lower factorial is the difference of the indices.

Example 3. Express ¹⁰P₄ in terms of factorials.

The upper factorial is the upper index, and the lower factorial is the difference of the indices. When the 6!'s cancel, the numerator becomes 10· 9· 8· 7.

This is the number of permutations of 10 different things taken 4 at a time.

Example 4. Calculate ⁿP_n.

ⁿP_n is the number of permutations of n different things taken n at a time -- it is the total number of permutations of n things: n!. The definition 0! = 1 makes line (1) above valid for all values of k: k = 0, 1, 2, . . . , n.

Problem 1. Write down all the permutations of xyz.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click 'Refresh' ('Reload').

xyz, xzy, yxz, yzx, zxy, zyx.

Problem 2. How many permutations are there of the letters pqrs?

4! = 1· 2· 3· 4 = 24

Problem 3. a) How many different arrangements are there of the letters of the word numbers?

7! = 5,040

b) How many of those arrangements have b as the first letter?

Set b as the first letter, and permute the remaining 6. Therefore, there are 6! such arrangements.

c) How many have b as the last letter -- or in any specified position?

The same. 6!.

d) How many will have n, u, and m together?

Begin by permuting the 5 things -- num, b, e, r, s. They will have 5! permutations. But in each one of them, there are 3! rearrangements of num. Consequently, the total number of arrangements in which n, u, and m are together, is 3!· 5! = 6· 120 = 720.

Problem 4. a) How many different arrangements (permutations) are there of the digits 01234?

5! = 120

b) How many 5-digit numbers can you make of those digits, in which the

b) first digit is not 0, and no digit is repeated?

Since 0 cannot be first, remove it. Then there will be 4 ways to choose the first digit. Now replace 0. It will now be one of 4 remaining digits. Therefore, there will be 4 ways to fill the second spot, 3 ways to fill the third, and so on. The total number of 5-digit numbers, then, is 4· 4! = 4· 24 = 96.

c) How many 5-digit odd numbers can you make with 0, 1, 2, 3, 4, and
c) no digit is repeated?

Again, 0 cannot be first, so remove it. Since the number must be odd, it must end in either 1 or 3. Place 1, then, in the last position. _ _ _ _ 1. Therefore, for the first position, we may choose either 2, 3, or 4, so that there are 3 ways to choose the first digit. Now replace 0. Hence, there will be 3 ways to choose the second position, 2 ways to choose the third, and 1 way to choose the fourth. Therefore, the total number of odd numbers that end in 1, is 3· 3· 2· 1 = 18. The same analysis holds if we place 3 in the last position, so that the total number of odd numbers is 2· 18 = 36.

Problem 5.

a) If the five letters a, b, c, d, e are put into a hat, in how many different
a) ways could you draw one out? 5

b) When one of them has been drawn, in how many ways could you
a) draw a second? 4

c) Therefore, in how many ways could you draw two letters? 5· 4 = 20

This number is denoted by ⁵P₂.

d) What is the meaning of the symbol ⁵P₃?

The number of permutations of 5 different things taken 3 at a time.

e) Evaluate ⁵P₃. 5· 4· 3 = 60

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Problem 6. Evaluate

a) ⁶P₃= 120 b) ¹⁰P₂= 90

c) ⁷P₅= 2520

Problem 7. Express with factorials.

See Permutations with Some Identical Elements

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