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# PERMUTATIONS AND COMBINATIONS

The Fundamental Principle of Counting

Section 2

Combinations

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 will be 3 ways to choose the second. That is, to each of those possible 4 there will correspond 3. Therefore, there are 4 · 3  or 12  possible ways to choose two letters from four.

 ab ba ca da ac bc cb db ad bd cd dc

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!.  (Topic 19.)

(To say "taken 4 at a time" is a convention.  We mean, "4! is the number of permutations of all 4 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 things 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."

 8P3 = 8·7·6 = 56·6 = 50·6  +  6·6 = 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,

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

Factorial representation

We saw in the Topic on factorials,

 8!5! = 8·7·6

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 follows:

 8P3 = 8!    (8 − 3)! = 8!5!

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

 nPk = n!    (n − k)! .   .   .   .   .   .   .(1)

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

Example 3.   Express  10P4  in terms of factorials.

 Solution. 10P4 = 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 fraction reduces to 10·9·8·7.

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

Example 4.   Calculate  nPn.

 Solution.    nPn = n!    (n − n)! = n!0! = n!1 = n!

nPn  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.

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.   How many different arrangements (permutations) are there of the digits 34567?

5! = 1·2·3·4·5 = 120

Problem 4.

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)   After one of them has been drawn our, 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?

It is the number of permutations of 5 different things taken 3 at a time.

e)   Evaluate 5P3.    5·4·3 = 60

Problem 5.   Evaluate

a)   6P3  = 120             b)   10P2  = 90

c)   7P5  = 2520

Problem 6.   Express with factorials.

 a)   nPk n!    (n − k)! b)   12P7 12! 5! c)   8P2 8!6! d)   mP0 m!m!

Problem 7.   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 8.   a)  How many permutations are there of the digits 01234?

5! = 120

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

b)  the 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.

Section 2:  Combinations Please make a donation to keep TheMathPage online.
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