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PERMUTATIONS Permutations are similar to combinations but extend the requirements
of combinations by considering order.
Figure 81."Tree" diagram. Suppose
we have two letters, A and B, and wish to know how many arrangements
of these letters can be made. Obviously the answer is two; that is, AB and BA If
we extend this to the three letters A, B, and
C, we find the answer to be ABC, ACB, BAC, BCA, CAB, CBA We
had three choices for the first letter; after we chose the first letter, we had
only two choices for the second letter; and after the second letter, we had only one choice. This
is shown in the "tree" diagram in figure 8l. Notice that a total of
six different paths lead to the ends of the "branches" of the
"tree" diagram. If
the number of objects is large, the tree would become very complicated;
therefore, we approach the problem in another manner, using parentheses to
show the possible choices. Suppose we were to arrange six objects in as many
different orders as possible. For the first choice we have six objects: STARTING POINT (6)(
)( )( )( )( ) For
the second choice we have only five choices: (6)(5)(
)( )( )( ) For
the third choice we have only four choices: (6)(5)(4)(
)( )( ) This
may be continued as follows: (6)(5)(4)(3)(2)(1) By
applying the principle of choice, we find the total possible ways of arranging
the objects to be the product of the individiual choices; that is, 6^{.}5^{.}4^{.}3^{.}2^{.}1 and
this may be written as 6! This leads to the statement: The
number of permutations of n objects taken all together is equal to n!. EXAMPLE: How many permutations of seven
different letters may be made? SOLUTION: We could use a "tree" diagram, but this
would become complicated. (Try it to see why.) We could use the parentheses as
follows: (7)(6)(5)(4)(3)(2)(1) = 5040 The easiest solution is to use the previous statement and
write
that is, the number of possible arrangements of seven
objects taken seven at a time is 7^{1}. NOTE: Compare this with the number of combinations of
seven objects taken seven at a time. If we are faced with finding the number of permutations of
seven objects taken three at a time, we use three parentheses as follows: In the first position we have a choice of seven objects: In the second position we have a choice of six objects: (7)(6)( ) In the last position we have a choice of five objects: (7)(6)(5) Therefore by principle of choice, the solution is 7^{.}6^{.}5 = 210 At this point we will use our knowledge of combinations to
develop a formula for the number of permutations of n objects taken r at a
time. Suppose we wish to find the number of permutations of five
things taken three at a time. We first determine the number of groups of three
as follows:
Thus, we have 10 groups of 3 objects each. We are now asked to arrange each of these 10 groups in as
many orders as possible. We know that the number of permutations of three
objects taken together is 3!. We may arrange each of the 10 groups in 3! or 6
ways. The total number of possible permutations of _{5}C_{3 }then
is _{
} which can be written as
The corresponding general form is
Knowing that
then _{
} But
therefore,
This formula is read: The number of permutations of n objects taken r at a
time is equal to n factorial divided by n minus r factorial. EXAMPLE: How many permutation of six objects taken two at a time
can be made? SOLUTION:
The number of permutations of six objects taken two at a time is written
EXAMPLE: In how many ways can eight people be arranged in a row? SOLUTION:
All eight people must be in the row; therefore, we want the number of
permutations of eight people taken eight at a time, which is
(remember
that 0! was defined as equal to 1) then
Problems
dealing with combinations and permutations often require the use of both
formulas to solve one problem. EXAMPLE: Eight first class and six second class petty officers are on the board
of the 56 club. In how many ways can
the members elect, from the board, a president, a vicepresident, a secretary,
and a treasurer if the president and secretary must be first class petty
officers and the vicepresident and treasurer must be second class petty
officers? SOLUTION:
Since two of the eight first class petty officers are to fill two different
offices, we write
Then,
two of the six second class petty officers are to fill two different offices;
thus, we write
The
principle of choice holds in this case; therefore, the members have 56^{.}30
= 1,680 ways
to select the required office holders. EXAMPLE: For the preceding example, suppose we are asked the following: In how
many ways can the members elect the office holders from the board if two of the
office holders must be first class petty officers and two of the office holders
must be second class petty officers? SOLUTION: We have already determined how many ways eight things may be taken two
at a time, how many ways six things may be taken two at a time, and how many
ways they may be taken together; that is, _{
} and _{
} then _{
} Our problem now is to find how many ways the members can
combine the four offices two at a time. Therefore, we write
Then, in answer to the problem, we write
In other words, if the members have _{
}ways of combining the four offices and
then for every one of these ways, the members have _{
}ways of arranging the office holders,
then they have
ways of electing the petty officers. 
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