Discrete

Mathematics

Counting

1
The Basics of counting
 A counting problem:
Each user on a computer system has a
password, which is six to eight characters long,
where each characters is an uppercase letter or a
digit. Each password must contain at least one digit.
How many possible passwords are there?

 This section introduces

a variety of other counting problems
the basic techniques of counting.

2
Basic counting principles
The sum rule:
If a first task can be done in n1 ways and
a second task in n2 ways, and if these tasks n1 n2
cannot be done at the same time. then there
are n1+n2 ways to do either task.
n1 + n2 ways

Example 1
 Suppose that either a member of the mathematics faculty or
a student who is a mathematics major is chosen as a
representative to a university committee. How many different
choices are there for this representative if there are 37
members of the mathematics faculty and 83 mathematics
majors and no one is both a faculty member and a student?

3
Sol: There are 37 ways to choose a member of the
mathematics faculty and there are 83 ways to choose a student
who is a mathematics major. Choosing a member of the
mathematics faculty is never the same as choosing a student
who is a mathematics major because no one is both a faculty
member and a student.

By the sum rule it follows that there are 37 + 83 = 120

possible ways to pick this representative.

4
Example 2 A student can choose a computer
project from one of three lists. The three lists contain
23, 15 and 19 possible projects respectively.
How many possible projects are there to choose
from?

Sol: 23+15+19=57 projects.

The product rule: n1

Suppose that a procedure can be broken down n1 × n2
ways
into two tasks. If there are n1 ways to do the n2
first task and n2 ways to do the second task after
the first task has been done, then there are
n1 n2 ways to do the procedure.
5
Example 3 The chair of an auditorium (The great Hall) is
　 to be labeled with a letter and a positive integer
　 not exceeding 100. What is the largest number of
　 chairs that can be labeled differently?

Sol: 26 × 100 = 2600 ways to label chairs.

letter 1  x  100
x Ν

Example 4 How many different bit strings are

there of length seven?
Sol: 　 1 　 2 　 3 　 4 　 5 　 6 　 7
□ □ □ □ □ □ □ → 27 ways
↑ ↑ ↑ ↑ ↑ ↑ ↑
0,1 0,1 0,1 . . . . . . 0,1

6
Example 4
How many different license plates are available if each
plate contains a sequence of 3 letters followed by 3
digits ?
Sol: □ □ □ □ □ □ →263 ． 103

letter digit

Example 6 How many functions are there from a

set with m elements to one with n elements?
Sol: f(a1)=? Can be b1 ～ bn, Total n
kinds a1 f b1
a2 b2
f(a2)=? Can be b1 ～ bn, Total n
． ．
kinds ． ．

． ． ：
am bn f(am)=? Can be b1 ～ bn, Total n
7
Example 5 How many one-to-one functions are there
from a set with m elements to one with n element?
(m  n)

Sol: f(a1) = ? Can be b1 ～ bn, a total of n functions

f(a2) = ? Can be b1 ～ bn, but not = f(a1), a total of n1
functions
f(a3) = ? Can be b1 ～ bn, but not = f(a1), or =f(a2),
a total of n2 functions
：
：
f(am) = ? Not f(a1), f(a2), ... , f(am1), a total of n(m1)
functions
∴there are a total of n ． (n1) ． (n2) ． ... ．
(nm+1) 1-1 functions #
8
Example 6 Each user on a computer system has a
password which is 6 to 8 characters long, where each
character is an uppercase letter or a digit. Each
password must contain at least one digit.
How many possible passwords are there?

Sol: Pi : # of possible passwords of length i , i=6,7,8

P6 = 366 266
P7 = 367 267
P8 = 368 268

∴ P6 + P7 + P8 = 366 + 367 + 368  266  267 268

possible passwords

9
Example 7
In a version of Basic, the name of a variable
is a string of one or two alphanumeric characters, where
uppercase and lowercase letters are not distinguished.
Moreover, a variable name must begin with a letter and
must be different from the five strings of two characters
that are reserved for programming use. How many
different variable names are there in this version of
Basic?

Sol:
Let Vi be the number of variable names of length
i.
V1 =26
V2 =26 ． 36 – 5
∴26 + 26 ． 36 – 5 different names. 10
※ The Inclusion-Exclusion Principle

A B  A  B  A B
A B

Example 8 How many bit strings of length eight either

start with a 1 bit or end with the two bits 00 ?
Sol:
１２３４５６７８
□□□□□□□□
↑ ↑ ......
① 1 0,1 0,1 → A total of 27 bit strings
② 　 . . . . . . . . . . . . 0 0 → A total of 26 bit strings
③ 1 . . . . . . . . . . . 0 0 → A total of 25 bit strings
 27 +26 25 bit strings 11
Example 9 A computer company receives 350
applications from computer graduates for a job planning
a line of new Web servers. Suppose that 220 of these
applicants majored in computer science, 147 majored in
business, and 51 majored both in computer science and
in business. How many of these applicants majored
neither in computer science nor in business?

Sol:

34 of the applicants majored neither in computer science nor in business

12
 The subtraction rule, or the principle of
inclusion–exclusion, can be generalized to find
the number of ways to do one of n different tasks
or, equivalently, to find the number of elements
in the union of n sets, whenever n is a positive
integer.
 For 3 sets
|A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|
Example 10
A total of 1232 students have taken a course in Spanish, 879 have taken a
course in French, and 114 have taken a course in Russian. Further, 103 have
taken courses in both Spanish and French, 23 have taken courses in both
Spanish and Russian, and 14 have taken courses in both French and
Russian. If 2092 students have taken at least one of Spanish, French, and
Russian, how many students have taken a course in all three languages?
Sol:
|S ∩ F ∩ R| = 7.
13
 The Principle of Inclusion–Exclusion
 Let A1,A2, . . . , An be finite sets. Then

 The inclusion–exclusion principle gives a formula for the

number of elements in the union of n sets for every
positive integer n. There are terms in this formula for the
number of elements in the intersection of every
nonempty subset of the collection of the n sets is.
Hence, there are 2 n  1 terms in this formula.
Example 11
Give a formula for the number of elements in the union of
four sets.
14
※ Tree Diagrams
Example10 How many bit strings of length four do
not have two consecutive 1s ?
Sol:
0 (0000)
0
0 1 (0001)
0 0 (0010) ∴ 8 bit strings
1
0 (0100)
1 0 1 (0101)
0 0 0 (1000)
1
1 (1001)
1 0 (1010)

bit 1 bit 3

15
 Exercise
How many bit strings of length ten both begin and end with a 1?
How many strings of five ASCII characters contain the character @ (“at” sign) at
least once? [Note: There are 128 different ASCII characters.
How many strings of three decimal digits

a) do not contain the same digit three times?

b) begin with an odd digit?
c) have exactly two digits that are 4s?
How many license plates can be made using either two uppercase English letters
followed by four digits or two digits followed by four uppercase English letters?
How many subsets of a set with 100 elements have more than one element?
A palindrome is a string whose reversal is identical to the string. How many bit
strings of length n are palindromes?
How many positive integers not exceeding 100 are divisible either by 4 or by 6?
How many ways are there to arrange the letters a, b, c, and d such that a is not
followed immediately by b?

16
Ex 40. How many subsets of a set with 100 elements
have more than one element ?
Sol: 100  100  100  100 
(1)          ...       2100  101
100   99   98   2 
( 2)  a1 , a2 ,..., a100  Thm. 4 of §4.3

An empty collection and only

Placed or Placed or Placed or
1 individual elements of the
Not placed Not placed Not placed
collection

Ex 41. A palindrome is a string whose reversal

is identical to the string. How many bit strings
of length n are palindromes ?
( abcdcba is a palindrome, abcd is not )

Sol: If a1a2 ... an is a palindrome, then

a1=an, a2=an1, a3=an2, … 17
The Pigeonhole Principle

Theorem 1 (The Pigeonhole Principle)

If k+1 or more objects are placed into k boxes, then
there is at least one box containing two or more of the
objects.

Proof
Suppose that none of the k boxes contains more than
one object. Then the total number of objects would
be at most k. This is a contradiction.

Example 1. Among any 367 people, there must be

at least two with the same birthday, because there
are only 366 possible birthdays.
18
Example 2 In any group of 27 English words, there must
be at least two that begin with the same letter.

Example 3 How many students must be in a class to

guarantee that at least two students receive the
same score on the final exam ? (0~100 points)

Sol: 102. (101+1)

Theorem 2. (The generalized pigeon hole principle)

If N objects are placed into k boxes, then
N
there is at least one box containing at least  k 
objects.

e.g. 21 objects, 10 boxes  there must be one box

containing at least  21  3 objects.
10  19
 100 
Example 4 Among 100 people there are at least  12   9
 
who were born in the same month. ( 100 objects, 12
boxes)

20
Example 5: During a month with 30 days a baseball
team plays at least 1 game a day, but no more
than 45 games. Show that there must be a period
of some number of consecutive days during which the
team must play exactly 14 games.

day 1 2 3 4 5 ... 15 30
# of game 3 2 1 2 sum  45
Number of games for some time =14

21
Sol:
Let aj be the number of games played on or before the
jth day of the month. (1  j  30)

Then a1 , a2 ,..., a30 is an increasing sequence of distinct

integers with 1  a j  45 j
(i.e., 1  a1  a2  a3  ...  a30  45)

Moreover, a1  14, a2  14,..., a30  14 is also an increasing

sequence of distinct integers with 15  a j  14  59
(i.e., 15  a1  14  a2  14  a3  14  ...  a30  14  59)

There are 60 positive integers a1 ,..., a30 , a1  14,..., a30  14

between 1 and 59. Hence,  i and j such that
ai  a j  14 (i.e., exactly 14 games were played
from day j  1 to day i )# 22
Def. Suppose that a1 , a2 ,..., a N is a sequence of numbers.
A subsequence of this sequence is a sequence of
the form ai1 , ai2 ,..., aim where 1  i1  i2  ...  im  N
(i.e., A subsequence is a sequence obtained from the original sequence)

e.g. sequence: 8, 11, 9, 1, 4, 6, 12, 10, 5, 7

subsequence: 8, 9, 12 ()
9, 11, 4, 6 ()

Def. A sequence is called increasing if ai  ai 1

A sequence is called decreasing if ai  ai 1
A sequence is called strictly increasing if ai  ai 1
A sequence is called strictly decreasing if ai  ai 1

23
Theorem 3. Every sequence of n2+1 distinct real
numbers contains a subsequence of length n+1 that is
either strictly increasing or strictly decreasing.

Example 6. The sequence 8, 11, 9, 1, 4, 6, 12, 10, 5,

7 contains 10=32+1 terms (i.e., n=3).
There is a strictly increasing subsequence of length
four, namely, 1, 4, 5, 7. There is also a decreasing
subsequence of length 4, namely, 11, 9, 6, 5.

Exercise 7 Construct a sequence of 16 positive

integers that has no increasing or decreasing
subsequence of 5 terms.
Sol: 4,3,2,1 8,7,6,5 12,11,10,9 16,15,14,13

24
 Exercise
Show that among any group of five (not necessarily consecutive)
integers, there are two with the same remainder when divided by 4.


a) Show that if five integers are selected from the first eight
positive integers, there must be a pair of these integers
with a sum equal to 9.
b) Is the conclusion in part (a) true if four integers are selected
rather than five?
a)How many numbers must be selected from the set { I , 2, 3, 4, 5, 6}
to guarantee that at least one pair of these numbers add up to 7?
b)There are 38 different time periods during which classes at a
university can be scheduled. If there are 677 different classes, how
many different rooms will be needed?

25
Permutations and Combinations

Def. A permutation of a set of distinct objects is an ordered

arrangement of these objects. An ordered arrangement of r
elements of a set is called an r-permutation.

Example 1. Let S = {1, 2, 3}.

The arrangement 3,1,2 is a permutation of S.
The arrangement 3,2 is a 2-permutation of S.

Theorem 1. The number of r-permutation of a set with n

n!
distinct elements is P ( n, r )  n  ( n  1)  ( n  2)...( n  r  1) 
position ： 1 2 3 … r ( n  r )!
　　　 □ □ □ … □
　　　
insert ： n n  1 n  2 … n  r  1 26
Example 2. How many different ways are there to select
a first-prize winner, a second-prize winner, and a
third-prize winner from 100 different people who have
entered a contest ?
Sol:
P (100,3)  100  99  98

Example 3. Suppose that a saleswoman has to visit

8 different cities. She must begin her trip in a specified
city, but she can visit the other cities in any order she
wishes. How many possible orders can the saleswoman
use when visiting these cities ?
Sol:
7! 5040

27
Def. An r-combination of elements of a set is an
unordered selection of r elements from the set.

Example 4: Let S be the set {1, 2, 3, 4}.

Then {1, 3, 4} is a 3-combination from S.

Theorem 2 The number of r-combinations of a set

with n elements, where n is a positive integer and r is
an integer with 0  r  n, equals
p ( n ,r ) n!
C  C ( n, r ) 
n
r ( nr )  r!
 r!( n  r )!
Known as the binomial coefficient
pf :
P (n, r )  C (n, r )  r!

28
Example 10. We see that C(4,2)=6, since the
2-combinations of {a,b,c,d} are the six subsets
{a,b}, {a,c}, {a,d}, {b,c}, {b,d} and {c,d}

Corollary 2. Let n and r be nonnegative integers with r  n.

Then C(n,r) = C(n,nr)

pf : From Thm 2.
n! n!
C ( n, r )    C ( n, n  r )
r!(n  r )! (n  r )!(n  (n  r ))!

Meaning: Selecting r elements from n elements, is

equivalent to taking out r elements leaving n-r elements
remaining

29
Example 6. How many ways are there to select 5
players from a 10-member tennis team to make a trip to
a match at another school ?
Sol: C(10,5)=252

Example 7. Suppose there are 9 faculty members in

the math department and 11 in the computer science
department. How many ways are there to select a
committee if the committee is to consist of 3 faculty
members from the math department and 4 from the
computer science department?

Sol: C (9,3)  C (11,4)

30
 Exercise
How many permutations of {a , b, e, d, e , t, g} end with a?
How many bit strings of length 10 contain
¨ exactly four 1s?
¨ at most four 1s?
¨ at least four 1s?
¨ an equal number of 0s and 1s?
a)A group contains n men and n women. How many ways are there to arrange these
people in a row if the men and women alternate?
b)How many permutations of the letters ABCDEFG contain
a) the string BCD? b) the string CFGA?
c) the strings BA and GF? d) the strings ABC and DE?
e) the strings ABC and CDE? f) the strings CBA and BED?
a)Suppose that a department contains 1 0 men and 1 5 women. How many ways
are there to form a committee with six members if it must have the same number of
men and women?
b)Suppose that a department contains 1 0 men and 1 5 women. How many ways
are there to form a committee with six members if it must have more women than
men?
31
 Binomial Coefficients
Example 1.
( x  y ) 3  ( x  y )( x  y )( x  y )  ? x 3  ? x 2 y  ? x y 2  y 3
To obtain a term of the form xy2 ，
a y must be chosen from two of the brackets, and an x
must be chosen from one of the
brackets
3

(Note ： x and y in the same bracket cannot be multiplied) 2 ) xy 2
The coefficient = (
( x  yare
∴There ) 3 different
( 30 ) x 3  (ways
3
1 ) x y 
2 to get
( 3
2 ) xxy
y 22
 ( 3
3 ) y 3

∴
Theorem 1. (The Binomial Theorem)
Let x,y be variables, and let n be a positive integer, then
n
( x  y ) n  ( 0n ) x n  (1n ) x n 1 y  ...  ( nn 1 ) xy n 1  ( nn ) y n   ( nj ) x n  j y j
j 0

32
Example 2. What is the coefficient of x12y13 in
the expansion of (2 x  3 y ) ?
25

Sol: (2 x  3 y ) 25  (2 x  ( 3 y )) 25

∴ 25
(13 )  212  (3)13

Cor 1. Let n be a positive integer. Then

n

k 0 1
( n

k 0
)  ( n
)  ( n
)  ...  ( n
n )  2 n

pf : By Thm 1, let x = y = 1
(1  1) n  ( 0n )  (1n )  ( n2 )  ...  ( nn )
n

Let n be a positive integer. Then   (k )  0

k n
Cor 2. ( 1)
k 0

pf : by Thm 1. (11)n = 0
33
Theorem 2. (Pascal’s identity)
Let n and k be positive integers with n  k
Then
 n  1  n   n 
       
 k   k  1  k 

PASCAL’s triangle
( 00 ) 1
(10 ) (11 ) 1 1
2
1 2 1
2 2
( ) ( ) ( )
0 1 2 1 3 3 1
( 30 ) (13 ) ( 32 ) ( 33 ) 1 4 6 4 1

…
…

( 34 )

34
  n  1  n   n 
       
pf : ①(algebraic proof)  k   k  1  k 

 n  1  (n  1)!
 k 
  k!( n  1  k )!
 n   n  n! n!
 k  1  k  
    (k  1)!(n  k  1)! k!(n  k )!
k  n! (n  k  1)  n! (n  1)  n!
  
k!(n  k  1)! k!(n  k  1)! k!( n  k  1)!

②(combinatorial proof):
n 1 n 1 n 1
‧ = ‧ + ‧  n  1   n  1    n 1
  k   k  1 0   k 1
       
k k 0 k 1
1 35
 Theorem 3. (Vandermode’s Identity)
m, n, r    , 0  r  m, n
r
C (m  n, r )   C (m, r  k )  C (n, k )
k 0

pf :
m n
mn mn mn
C(m+n, r) = = ↓ ↓ + ↓ ↓ +...+ ↓ ↓
r, 0 r1, 0 0, r
r
  m  n    m  n   ...   m  n 
 r  0   r  1 1   0  r 
r
  C (m, r  k )  C (n, k )
k 0
36
37
Generalized Permutations and Combinations
Permutations with Repetition
Example 1 How many strings of length r can be formed
from the English alphabet?
Sol. 26r
Thm 1. The number of r-permutations of a set of n objects
with repetition allowed is n r.

38
Thm 2. There are C(r+n1, r) r-combinations from
a set with n elements when repetition of elements is
allowed.
pf : (Each r-combination of a set with n elements when
repetition is allowed can be represented by a list of n − 1 bars
and r stars. The n − 1 bars are used to mark off n different
cells, with the ith cell containing a star for each time the ith
element of the set occurs in the combination)

Example ： n = 4, set is {a1, a2, a3, a4}, r = 6

A2 appear 1 time a3 does not appear

＊＊｜＊｜｜＊＊＊ {a1 , a1 , a2 , a4 , a4 , a4 }
a1 appears 2 a4 appears 3 times ∴  r  n  1 combinations #
times  r 
39
Example 3. Suppose that a cookie shop has 4 different
kinds of cookies. How many different ways can
6 cookies be chosen?

Sol: The number of ways to choose six cookies is the number

of 6-combinations of a set with four elements. From Theorem 2
this equals C(4 + 6 − 1, 6) = C(9, 6) ways

Example 4. How many solutions does the equation

x1 + x2 + x3 = 11
have, where x1, x2, x3 are nonnegative integers?’

Sol: 11 stars to be inserted between 2 bars

11  2 
By theorem 2  11  solutions
 
40
※ since x1, x2, x3 are nonnegative integers, x1  1, x2  2, x3  3,
Original equation is x1 + x2 + x3 = 11
can be changed to (x1 1) + (x22) + (x33) = 11123 = 5
At this point y1 + y2 + y3 = 5
Where y1 = x1 1, y2 = x2 2, y3 = x3 3 and y1, y52,2y3 N
∴5 stars to be inserted between 2 bars   5 
 
solutions
(Note ： case y1 = 1, y2 = 3, y3 = 1 is equal to x1 =2, x2 =5, x3 =4)

※if the problem is 1 x1  3, x2  2, x3  3 ， you need to exclude

x1 > 3 case
(i.e x1  4 case)
because (x1 4) + (x22) + (x33) = 119 = 2

∴ A total of  5 5 2    2 2 2  combinations
    41
42
 ※ Permutations with indistinguishable objects
Example 5. How many different strings can be made
by reordering the letters of the word SUCCESS ?
Sol:
There are 3 Ss, 2 Cs, 1 U and 1 E,
The three Ss can be placed among the seven positions in C(7,3)
different ways leaving four positions free.
Then the two Cs can be placed in C(4, 2) ways, leaving two free
positions.
Then the two Cs can be placed in C(2, 1) ways, leaving two free
positions.
Hence E can be placed in C(1, 1) way
Consequently, from the product rule, the number of different
strings that can be made is
C(7, 3)C(4, 2)C(2, 1)C(1, 1) = 420 43
Thm 3. The number of different permutations of n objects,
where type 1 ： n1 ways
n!
type 2 ： n2 ways is
n1!n2! nk !

type k ： nk ways
pf :
 n  n  n1  n  n1  n2   n  n1  n2  　　  nk 1  n!
      
n n
 1  2  n 3   nk  n1!n2 ! 　 nk !

44
 Exercise
Solve the following
a) How many ways are there to deal hands of five cards to six players
from a standard 52-card deck?
b) How many ways are there to distribute n distinguishable objects
into k distinguishable boxes so that ni objects are placed in box i?
How many ways are there to choose a dozen apples from a bushel
containing 20 indistinguishable Delicious apples, 20 indistinguishable
Macintosh apples, and 20 indistinguishable Granny Smith apples, if at
least three of each kind must be chosen?
How many ways are there to pack eight identical DVDs into five
indistinguishable boxes so that each box contains at least one DVD?
How many ways are there to distribute six indistinguishable objects into
four indistinguishable boxes so that each of the boxes contains at least
one object?

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Distributing objects into Boxes
Distinguishable Objects and distinguishable boxes
Example 8. How many ways are there to distribute
hands of 5 cards to each of four players from the
standard deck of 52 cards?
 52 
Sol: player 1 ：  5 
ways
 47 
player 2 ： from the remaining 5  5 
52! 47! 42! 37! 52!
  52  47  42  37      
 5  5  5  5  5! 47! 5! 42! 5! 37! 5! 32! 5! 5! 5! 5! 32!
Note: the above question is equivalent to 52 cards in 5 different box
method,
that is box 1 for player 1,
　　　　 box 2 for player 2
　　　　 box 3 for player 3
　　　　 box 4 for player 4 46
　 while box 5 is for the cards left.
 Exercise
How many ways can n books be placed on k
distinguishable shelves
a) if the books are indistinguishable copies of
the same title?
b) if no two books are the same, and the
positions of the books on the shelves matter?

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Thm 4. The number of ways to distribute n distinguishable
objects into k distinguishable boxes so that ni
objects are placed into box i, i=1, 2, …, k, equals
n!
n1! n2 ! nk ! (the same as Thm 3)

Indistinguishable Objects and distinguishable boxes

Example 9. How many ways are there to place 10

indistinguishable balls into eight distinguishable bins?
Sol: C(8+101, 10)
There are C(n+r1, n1) ways to place r indistinguishable
objects into n distinguishable boxes.
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Distinguishable Objects and indistinguishable boxes
Example 6. How many ways are there to put four different
employees into three indistinguishable offices, when each
office can contain any number of employees?
Sol: employees: A, B, C, D
4 people in the same office: {{A, B, C, D}}  1 way
3 people in the same office, 1 person in another office
: {{A, B, C}, {D}}, {{A, B, D}, {C}},… 4 ways
2 people in the same office, 2 people in another office
: {{A, B}, {C, D}}, {{A, D}, {B, C}},… 3 ways
2 people in the same office, The other 2 people in the remaining two Office:
{{A, B}, {C}, {D}}, {{A, D}, {B}, {C}}, …  6 ways
Total: 14 ways
Note. There is no simple closed formula. You may refer to
Stirling numbers of the second kind. (p. 378)
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Indistinguishable Objects and indistinguishable boxes
Example 7. How many ways are there to pack six copies of
the same book into four identical boxes, where a box can
obtain as many as six books?
Sol: 3, 3 2, 2, 1, 1
6
5, 1 3, 2, 1
Total 9
4, 2 3, 1, 1, 1
4, 1, 1 2, 2, 2

Note. This problem is the same as writing n as the sum of at

most k positive integers in nonincreasing order.
That is, a1+a2+…+aj = n, where a1, a2, …, aj are positive
integers with a1  a2  …  aj and j  k.
No simple closed formula exists.
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