Lesson Permutations

Algebra ->  Permutations -> Lesson Permutations      Log On


   


This Lesson (Permutations) was created by by Alwayscheerful(414) About Me : View Source, Show
About Alwayscheerful: I'm available as an online paid tutor if anyone need extensive 1 on 1 help. =)

By the permutations of the letters abc we mean all of their possible arrangements:
abc
acb
bac
bca
cab
cba

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

This number was 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%2An.

For example, imagine the letters a, b, c, d put into a hat, and let us draw 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. Therefore, there are
4%2A3 or 12 possible ways to choose two letters from four.
ab
ba
ca
da
ac
bc
cb
db
ad
bd
cd
dc

b 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.

That is, the number of permutations of 4 different things taken 4 at a time is 4!.

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%21=5%2A4%2A3%2A2%2A1=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?

Answer:a) Let r be the second letter. Then there are 5 ways to fill the first spot, 4 ways to fill the third, 3 to fill the fourth,
and so on. There are 5! such permutations. 5%21=120.

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.


We have seen that the number of ways of choosing 2 letters from 4 is 4· 3 = 12. This is called
The number of permutations of 4
different things taken 2 at a time.

We write this as 4P%5B2%5D.

The lower index 2 indicates the number of factors. The upper index 4 indicates the first factor.
For example, 8P%5B3%5D means "the number of permutations of 8 different things taken 3 at a time." And 8P%5B3%5D
=8%2A7%2A6

=56%2A6

=50%2A6%2B6%2A6

=336.

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%5Bk%5D = n%2A%28n-1%29%2A%28n-2%29+%2A+ellipsis+%2A+%28n-k%2B1%29     (k factors)


Factorial representation
%288%21%29%2F%285%21%29=8%2A7%2A6

5%21 is a factor of 8%21, and therefore the 5%21 cancel.

Now, 8%2A7%2A6=8P%5B3%5D. We see, then, that 8P%5B3%5D can be expressed in terms of factorials as
8P%5B3%5D=8%21%2F%288-3%29%21=%288%21%29%2F%285%21%29


In general, the number of arrangements -- permutations -- of n things take k at a time, can be represented as follows:
nP%5Bk%5D=n%21%2F%28n-k%29%21

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



EXAMPLE 3

Express 10P%5B4%5D in terms of factorials.

Answer:10!/6!

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%2A9%2A8%2A7.
This is the number of permutations of 10 different things taken 4 at a time.



EXAMPLE 4

Calculate nP%5Bn%5D.

Answer:n%21%2F%28n-n%29%21%29= n%21%2F0%21 = n%21%2F1 = n%21.

Hope this helps!

This lesson has been accessed 25518 times.