Permute the Order in Various Objects

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= 2 Calculate our permutation value n P r for n = 5 and r = 3: 5 P 3 = 120: 2: 5 P 3 = 60. In Microsoft Excel or Google Sheets, you write this function as =PERMUT(5,3) View the Permutations and Combinations Flashcards Watch the Permutations. Permutation Problem 2. Choose 3 contestants from group of 12 contestants. At a high school track meet the 400 meter race has 12 contestants. The top 3 will receive points for their team.

Provides the generic function andmethods for permuting the order of various objectsincluding vectors, dendrograms (also hclust objects),the order of observations in a dist object, the rows and columns of a matrix, all dimensions of an array given a suitable ser_permutationobject.

Arguments

an object (a list, a vector, a dist object, a matrix, an array or any other object which provides dim and standard subsetting with '[').

an object of class ser_permutation which contains suitable permutation vectors for x.

additional arguments for the permutation function.

Details

The permutation vectors in ser_permutation are suitable if the number of permutation vectors matches the numberof dimensions of x and if the length of each permutation vectorhas the same length as the corresponding dimension of x.

For 1-dimensional/1-mode data (list, vector, dist), order can also be a single permutation vector of class ser_permutation_vector or data which can be automatically coerced to this class(e.g. a numeric vector).

For dendrograms and hclust, subtrees are rotated torepresent the order best possible. If the order is not achived perfectlythen the user is warned. This behavior can be changed with the extraparameter incompatible which can take the values 'warn' (default), 'stop' or 'ignore'.

See Also

ser_permutation,dist in package stats.

Aliases

• permute
• permute.dist
• permute.numeric
• permute.list
• permute.matrix
• permute.array
• permute.data.frame
• permute.hclust
• permute.dendrogram

Topics in

P R E C A L C U L U S

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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.

Permute 2 0 7

Example 3. Express ¹⁰P₄ in terms of factorials.

Permute 2 0 72

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

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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