Book 12 CBSE Notes Set Class 12

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M_XI_Sets 1 VMC/School Level Notes

TOPIC: SETS

A well-defined collection of distinct objects is called as set.

A set is a collection of well-defined objects. The term well-defined means that one should be able to clearly

specify that a particular element either belongs to or is excluded from the set (the collection).

For example:

(i) The collection of all English vowels is a set.

(ii) The collection of natural numbers is a set usually denoted by N.

Whereas, (i) collection of best movies ever made (ii) collection of dangerous animals in the world, are not sets

as the elements can be not well-defined.

Thus, set is not any collection but acollection of well-defined objects. Each object belonging to a set is

called an element of the set.

We generally use capital letters A, B, C, etc., to denote sets and lowercase letters a, b, c, x, etc., to denote

elements of a set.

The symbol is used to indicate belongs to (or is an element of ). Therefore, p is an element of S is

written as p S, and p is not an element of S (does not belong to) is written as pS.

e.g., Let V be the set of vowels in the English alphabet, then Va and Vd .

Representation of Sets

The following are the most common methods used for representating sets.

(i) Roster Method: A set may be described by listing all its elements. For example, the set V of vowels of the

English alphabet is: V = {a, e, i, o, u}.

(ii) Set-Builder Method: In this form, a set is described by means of a property that characterises all the

elements of the set (i.e. the property is shared by all the elements of the set and if a certain object has that

property, then it belongs to the set). A set S characterised by a property p of an object x may be written as

S { : ( )},x p x where p(x) means that x has property p.

(a) N = {n: n is a natural number} (b) P = {p: p is a prime number}

(c) E = {n: n is even number}.

Classification of Sets

Sets are classified as follows:

I. Empty Set : The set having no element is called empty set (or null set or void set) and is generally denotedby

(phi)or{}AND N OTBy{or{0}.

e.g., { :x x is a negative integer whose square is 1} =

II. Finite and Infinite Sets : A finite set is one that has finite (countable) number of elements, whereas an

infinite set is one that has infinite (uncountable) number of elements.

The number of elements in a set A is denoted byn (A) and is called the cardinality of the set.

e.g., A= {1, 2, 3, 4, 5} is the finite set and N {1,2,3,4,.......} is an example of an infinite set.

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All infinite sets cannot be described in the roster method. For example, a set of real numbers cannot be described

in this form because the elements of the set do not follow any particular pattern.

A set does not change if one or more elements of the set are repeated.

For example, A= {x : x all letter of the word WOLF} B={x : x all letters of the word FOLLOW}A=B={W,O,L,F}

III. Equal Sets: Two sets A and B are said to be equal if they have exactly same elements.

e.g., (i) Let A {1,2,3,7} and B {7,3,2,1} , then A = B

(ii) Let A { |x x is a prime number

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Subsets of Set of Real Numbers

There are many important subsets of R, the set of real numbers.

The set of irrational numbers, denoted by T, is composed of all other real numbers (other than Q).

Thus, T = { : Rx x and Q}x = R Q, i.e., all real numbers that are not rational.

Some of the obvious relation among these subsets is: N Z Q, Q R, T R, N T .

The set of natural numbers N = {1, 2, 3, 4, 5, }

The set of integers Z = {, 3, 2, 1, 0, 1, 2, 3, }

The set of rational numbers Q = { : , , Zpx x p qq

and 0}q

The set of irrational numbers, denoted by T, is composed of all other real numbers (other than Q).

Intervals as Subsets of R

Let a, bR and a < b. Then the set of real numbers {y : a < y < b} is called an open interval and is denoted by (a, b).

All the points between a and b belong to the open interval (a, b) but a, b themselves do not belong to this interval.

The interval that contains the end points also is called closed interval and is denoted by [a, b]. Thus,

[a, b]={ : }x a x b

We can also have intervals closed at one end and open at the other, i.e.

[a, b) = { : }x a x b is an open interval from a to b, including a but excluding b.

(a, b] = { : }x a x b is an open interval from a to b including b but excluding a.

The number (ba) is called the length of any of the interval (a, b), [a,b], [a,b) or (a,b].

Venn Diagram

Sets can be represented by diagrams known as Venn Diagrams.

We usually use a rectangle to represent a universal set U. Elements of U are denoted by dots inside the rectangle.

Elements of any given set U are enclosed in a circle or a closed region within the rectangle. When sets have a large

number of elements, the dots are omitted and only the closed region is shown.

Operations on Sets

Sets can be represented by diagrams known as Venn Diagrams.

We usually use a rectangle to represent a universal set U. Elements of U are denoted by dots inside the rectangle.

Elements of any given set U are enclosed in a circle or a closed region within the rectangle. When sets have a large

number of elements, the dots are omitted and only he closed region is shown.

There are number of different operations that can be performed on sets to get the new sets. Few such operations are listed

below.

Intersection of Two Sets: Let U denotes a universal set and let A and B be subsets of U. The set of elements that are in

both A and B is known as ther intersectionof A and B. We denote it as A B .

Thus, A B = intersection of A and B, = {all elements common to A and B}= { : Ax x and Bx }

The shaded regions in the given figure show the intersection of two sets A and B in each of the following cases.

(i) One of A or B is a proper subset of the other,

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(ii) None is a subset of the other but each contains some elements of the other, and

(iii) The two have no common elements.

Note that, in the third case, A B has no elements, i.e. A B = . Such sets are known as disjoint sets. Note further,

that in case (i) A B B, case (ii) A B A & B both and in case (iii) A B (= ) A or B.

Important Formulae

(i) Commutative Law: If A and B are two sets. Then, AB = BA.

(ii) Associative Law: If A, B and C are three sets, then, (AB)C = A (BC)

(iii) Law of and U: If A is any set, then, A A and U A A U A.

(iv) Idempotent Law: For any set A, A A A.

The law can be seen easily from the following Venn diagrams.

(i) (B C) (ii) A (B C) (iii) (B C)

(iv) (A C) (v) (A B) (A C)

Union of Two Sets

The totality of all the elements of two sets A and B is known as the Unionof A and B. It is denoted by A B . Thus,

A B is the set of all those elements that belong to A or to B or to both. The shaded region in the figure below shows

the union of A and B.

Note that A A B and B A B .

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If A = {1, 2, 3}, B = {4, 5, 1}, C = {p}. Then A B = {1, 2, 3, 4, 5} and A C = {1, 2, 3, p}.

Important Formulae

(i) Commutative Law: If A and B are two sets, then, A B = B A .

(ii) Associative Law: If A, B and C are three sets, then, (A B) C = A (B C)

(iii) Law of identity element, is the identity of U: if A is any set, then, A =A

(iv) Idempotent Law: For any set A, A A =A

(v) Law of U: U A=U

(vi) Distributive Law: distributes over

i.e. A (B C) (A B) (A C)

Similarly, distributes over

i.e. A (B C) (A B) (A C)

Difference of Sets: The difference of the sets A and B is the set of elements which belong to A but not to B. i.e.,

A B { : Ax x and B}x .

Similarly B A { : Bx x and A}x

Complement of a Set

Let U be the universal set and A U , then the complement of A is the set of all those elements of U which are not

elements of A. We denote the complement of A by A or ~A or (UA). The shaded region in the given Fig. represents

A , the complement of A.

Important Properties

1. Complement Law: (i) A A = U and (ii) A A =

2. De Morgans Law: (i) (A B) = A B and (ii) (A B) = A B

(iii) A B C A B A C (iv) A B C A B A C

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3. Law of double complementation: (A ) = A

4. Law of empty set and universal set(i) = U (ii) U =

If A is a subset of universal set U, then its complement A is also a subset of U.

Summery of theorems

If A, B and C are finite sets and U be the finite universal set, then

(i) A Bn A (B) A Bn n n

(ii) A Bn A (B) A and Bn n are disjoint non-void sets.

(iii) A Bn A A B . . A B A B An n i e n n n

(iv) Symmetric difference between 2 sets n(A B) = No. of elements which belong to exactly one of A or B

= n ((A B) (B A))

= n (A B) + n (B A) [ (A B) and (B A) are disjoint] = n (A) n (AB) + n (B) n (AB)

= n (A) + n (B) 2n (AB)

(v) n (ABC) = n (A) + n (B) + n (C) n (AB) n (BC) n (AC) + n (ABC)

(vi) No. of elements in exactly two of the sets A, B, C

= n (AB) + n (BC) + n (CA) 3 A B Cn

(vii) No. of elements in exactly one of the sets A, B, C

= n (A) + n (B) + n (C) 2 n (AB) 2n (BC) 2n (AC) + 3n (AB C)

(viii) n A B = n (A B) = n (U) n (AB)

(ix) n A B =n (A B) = n (U) n (AB)

(x) n A B = n (A) n (AB).

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PRACTICE QUESTIONS

1. Write the solution set of the equation x2 + x 2 = 0 in roster form.

2. Write the set A = {1, 4, 9, 16, 25, . . . } in set-builder form.

3. Describe set 1 1 1 1 1, , , ,4 8 10 32 64

in set -builder form.

4. Describe the following set using the set-selector (set-builder) method:

(i) A {2,4,6,8,10} (ii) B {5} (iii) C = {101, 102, 103, ..., 999}

[Ans. (i) : 2 1 5x x k k (ii) B : 5x x (iii) C : 3 all numbers / 100 1000x x b w and ] 5. Write the following intervals in set builder form:

(i) (5, 3) (ii) [7, 10] (iii) (5, 11], (iv) [20, 7 )

[Ans. (i) {x : Rx , 5 3x (ii) {x : Rx , 7 10x (iii) : R, 5 11x x x

(iv) : R, 20 7x x x ] 6. Write the following as intervals:

(i) {x: x R, 5

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15. Draw appropriate Venn diagram for each of the following: (i) (A B) (ii) A B (iii) (A B)

[Ans. 1. (i) (ii) (iii)

16. Write down the complement of the following sets. A = {x: x is a multiple of 3,} 7, Nx x and U = N

[Ans. A : is all N except 6x x ] 17. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ; A = {1, 2, 3, 4}; B = {4, 5, 6, 7}; C = {7, 8, 9, 10}

Find (i) A (ii) B (iii) C (iv) (A B) (v) (B C)

[Ans. (i) {5, 6, 7, 8, 9, 10} (ii) {1, 2, 3, 8, 9, 10} (iii) {1, 2, 3, 4, 5, 6} (iv) {4, 5, 6, ... 1} (v) {1, 2, 3, 7, 8 9, 10}. ]

18. If X and Y are two sets such that X Y has 50 elements, X has 28 elements and Y has 32 elements, how many

elements does X Y have.

19. In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach

both physics and mathematics. How many teach physics.

20. In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75

were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice

nor orange juice.

21. In a town 10,0000 families, it was found that 40% families buy newspaper A, 20% family by newspaper B and

10% buy newspaper C, 5% families buy A and B, 3% buy B, and C and 4% buy A and C. If 2% families buy all

the newspapers, find the no. of families which buy (i) A only (ii) B only (iii) one of A, B and C.

22. Each student in a class of 40, studies at least one of the subjects. Mathematics, English and Economics. 16 Study

English, 22 Economics and 26 Mathematics, 5 study English and Economics, 14 Mathematics and Economics and

2 English, Economics and Mathematics. Find the number of students who study: (i) English and Mathematics (ii)

English, Mathematics but not Economics.

23. A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58

men and only three men got medals in all the three sports, how many received medals in exactly two of the three

sports. [ Ans. 9]

24. There are 60 students in a Mathematics class and 90 students in a Physics class. Find the number of students which

are either in physics class or Mathematics class in the following cases :

(i) Two classes meet at the same hour. [Ans. 120]

(ii) Two classes meet at different hours and 30 students are enrolled in both the course. [Ans. 150]

25. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many

speak at least one of these two languages. [Ans. 60]

26. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100

were taking both tea and coffee. Find how many students were taking neither tea nor coffee. [Ans. 325]

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27. There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C1, 50 to chemical C2, and

30 to both the chemicals C1 and C2. Find the number of individuals exposed to

(i) Chemical C1 but not chemical C2 (ii) Chemical C2 but not chemical C1

(iii) Chemical C1 or chemical C2

28. A market research group conducted a survey of 1000 consumers and reported that 720 consumers like product A

and 450 consumers like product B, what is the least number that must have liked both products.

29. Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this

data correct.

30. A college warded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58

men and only three men got medals in all the three sports, how many received medals in exactly two of the three

sports.

31. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I,

9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find: (i) the number of

people who read at least one of the newspapers. (ii) the number of people who read exactly one newspaper.

32. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people

liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the

three products. Find how many liked product C only.

33. A market research group conducted a survey of 2000 consumers and reported that 1720 consumers like product A

and 1450 consumers like product B, what is the least number that must have liked both products. [Ans.1170]

34. In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student likes to play at least

one of the two games. How many students like to play both cricket and football. [Ans. 5]

35. Prove that (i) (A B)C = ACBC (ii) (A B)C = ACBC

36. Using properties of sets, show that: (i) A ( A B ) = A (ii) A ( A B ) = A.

37. Show that for any sets A and B, (i) A = ( A B ) ( A B ) (ii) A ( B A ) = ( A B )

38. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows

either Hindi or English. How many students are there in the group. [Ans.125]

39. Prove that: A (B C) = (A B) (A C)

EXPLANATION

1. Sol: The given equation can be written as (x 1) (x + 2) = 0, i.e., x = 1, 2.

Therefore, the solution set of the given equation can be written in roster form as {1, 2}.

2. Sol: We may write the set A as A = {x : x is the square of a natural number}

Alternatively, we can write A = {x : x = n2, where n N}

3. Sol: 1; , N, 2 62n

x x n n

7. Sol:A B = {2, 4} and B A = {7, 9}

8. Sol:(i) A = {3, 4, 7, 8, 9}, therefore A = set of all those elements of U which are not in A = {1, 2, 5, 6, 10}

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(ii) B = set of all those elements of U which are not in B = {4, 6, 7, 8, 9}

(iii) A = {3, 4, 7, 8, 9}; B = {1, 2, 3, 5, 10}; AB = {1, 2, 3, 4, 5, 7, 8, 9, 10}

(A B) = {set of those elements of U which are not in AB} = {6}.

(iv) A = {3, 4, 7, 8, 9}

C = {3, 5, 6, 8}

AC = {3, 4, 5, 6, 7, 8, 9}

(A C) = {1, 2, 10}

(v) A = {1, 2, 5, 6, 10}

(A ) = set of those elements of U are not in A = {3, 4, 7, 8, 9} = A

(vi) As shown above (A B) = {6}

A = {1, 2, 5, 6, 10}; B = {4, 6, 7, 8, 9}

A B = {6}

Hence (A B) = A B

B = {1, 2, 3, 5, 10}

AB = {1, 2, 3, 4, 5, 7, 8, 9, 10}

(A B) = {set of those elements of U which are not in AB} = {6}.

9. Sol:Let a A. Then a A B. Since A B = A B, a A B. So a B.

Therefore, A B. Similarly, if b B, then b A B. Since

A B = A B, b A B. So, b A. Therefore, B A. Thus, A = B.

10. Sol:Let X P ( A B ).

X A B. X A and X B.

X P ( A ) and X P ( B )

X P ( A ) P ( B).

This gives P (A B ) P ( A ) P ( B ). ........(i)

Now , Let Y P ( A ) P ( B ).

Y P (A) and Y P (B)

Y A and Y B.

Y A B Y P (A B).

This gives P (A) P (B) P (A B) .....(ii)

Hence P (A B ) = P ( A ) P ( B ).

11. Sol: B = {1, 2, 3, 5, 10}; C = {3, 5, 8, 6}

B C = set of those elements of B which are not in C = {1, 2, 10}

(B C)= set of those elements of U which are not in (B C) = {3, 4, 5, 6, 7, 8, 9}

12. Sol. By Venn Diagram

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By set-builder form

: :LHS x x A B C x x A and x B C .

:x x A and x B or x C :x x A and x B or x Aand x C

:x x A B or x A C :x x A B A C Proved

13. Sol. By the statement A = 5,9,13,17,21 ; B = 3,6,9,15,18,21,24

A B 5,13,17 , A B= 9, 21 , A B 3,5,6,9,13,15,17,18,21, 24

(i) A (A B) = {9, 21} (ii) 5,13,17A A B

(iii) 3,5,6,13,15,17,18,24A B A B

14. Sol. A B C is shown by the shaded region.

Then A B A C A B C

18. Sol: Given that: n (X Y) = 50, n (X) = 28, n (Y) = 32, n (X Y) = ?

By using the formula: n ( X Y ) = n ( X ) + n ( Y ) n ( X Y ),

We find that n ( X Y ) = n ( X ) + n ( Y ) n ( X Y ) = 28 + 32 50 = 10

19. Sol: Let M denote the set of teachers who teach mathematics and P denote the set of teachers who teach physics. In

the statement of the problem, the word or gives us a clue of union and the word and gives us a clue of

intersection. We, therefore, have

n ( M P ) = 20 , n ( M ) = 12 and n ( M P ) = 4

We wish to determine n ( P ).

Using the result, n ( M P ) = n ( M ) + n ( P ) n ( M P ),

We obtain, 20 = 12 + n ( P ) 4

Thus n ( P ) = 12

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Hence 12 teachers teach physics.

20. Sol: Let U denote the set of surveyed students and A denote the set of students taking apple juice and B denote the

set of students taking orange juice. Then

n (U) = 400, n (A) = 100, n (B) = 150 and n (A B) = 75.

Now n (A B) = n (A B) = n (U) n (A B) = n (U) n (A) n (B) + n (A B) = 400 100 150 + 75 =

225

Hence 225 students were taking neither apple juice nor orange juice.

21. Sol:

Given A, B, C as 40%, 20% , 10% resp.

n(A) = a + b + d + e = 4000; n(B) = b + c + e + f = 2000; n(C) = d + e + f + g = 1000

Also, A Bn = b + e = 5% of 10000 = 500

B Cn = e + f = 3% of 10000 = 300

A Cn = e + d = 4% of 10000 = 400.

A B C 2%n e of 10000 = 200 b = 300 ; f = 100; d = 200 and g = 500, c = 1400, a = 3300 (i) n (only A) = a = 3300

(ii) n (only B) = c = 1400

(iii) 10000 (a + b + c + d + e + f + g) = 10000 6000 = 4000

22. Sol: E = set of students who study English

M = Math

A = Economics

n E M A = 40, n(E) = 16, n(M) = 26, n(A) = 22, n E A = 5

14n M A , 2n M E A

(i) n A E M n A n E n M n A E n A M E M n A B C

40 = 22 + 16 + 26 5 14 2n E M 7n E M

(ii) 7 2 5n E M A n E M n E M A

27. Sol: Let U denote the universal set consisting of individuals suffering from the skin disorder, A denote the set of

individuals exposed to the chemical C1 and B denote the set of individuals exposed to the chemical C2.

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Here n (U) = 200, n (A) = 120, n (B) = 50 and n (A B) = 30

(i) From the Venn diagram given in Fig, we have

A = ( A B ) ( A B ).

n (A) = n( A B ) + n( A B ) (Since A B) and A B are disjoint.)

or n ( A B ) = n ( A ) n ( A B ) = 120 30 = 90

Hence, the number of individuals exposed to chemical C1 but not to chemical C2 is 90.

(ii) From the figure, we have B = ( B A) ( A B) and so, n (B) = n (B A) + n ( A B)

(Since B A and A B are disjoint.)

or n ( B A ) = n ( B ) n ( A B) = 50 30 = 20

Thus, the number of individuals exposed to chemical C2 and not to chemical C1 is 20.

(iii) The number of individuals exposed either to chemical C1 or to chemical C2, i.e.,

n ( A B ) = n ( A ) + n ( B ) n ( A B) = 120 + 50 30 = 140.

28. Sol: Let U be the set of consumers questioned, S be the set of consumers who liked the product A and T be the set

of consumers who like the product B. Given that

n ( U ) = 1000, n ( S ) = 720, n ( T ) = 450

So n ( S T ) = n ( S ) + n ( T ) n ( S T ) = 720 + 450 n (S T) = 1170 n ( S T )

Therefore, n ( S T ) is maximum when n ( S T ) is least. But S T U implies n ( S T ) n ( U ) = 1000.

So, maximum values of n ( S T ) is 1000. Thus, the least value of n ( S T ) is 170. Hence, the least number of

consumers who liked both products is 170.

29. Sol: Let U be the set of car owners investigated, M be the set of persons who owned car A and S be the set of

persons who owned car B.

Given that n ( U ) = 500, n (M ) = 400, n ( S ) = 200 and n ( S M ) = 50.

Then n ( S M ) = n ( S ) + n ( M ) n ( S M ) = 200 + 400 50 = 550

But S M U implies n ( S M ) n ( U ). This is a contradiction. So, the given data is incorrect.

30. Sol: Let F, B and C denote the set of men who received medals in football, basketball and cricket, respectively.

Then n ( F ) = 38, n ( B ) = 15, n ( C ) = 20

n (F B C ) = 58 and n (F B C ) = 3

Therefore, n (F B C ) = n ( F ) + n ( B ) + n ( C ) n (F B ) n (F C ) n (B C) + n ( F B C ),

gives n ( F B ) + n ( F C ) + n ( B C ) = 18

Consider the Venn diagram as given in figure.

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Here, a denotes the number of men who got medals in football and basketball only, b denotes the number of men

who got medals in football and cricket only, c denotes the number of men who got medals in basket ball and

cricket only and d denotes the number of men who got medal in all the three. Thus, d = n ( F B C ) = 3 and a +

d + b + d + c + d = 18.

Therefore a + b + c = 9, which is the number of people who got medals in exactly two of the three sports.

31. Sol:

e = 3

d + e = 9 6d

b + e = 11 8b

e + f = 8 5f

Also a + b + e + d = 25 8a

Similarly C = 10, g = 12

(i) People reading at least one news paper = 8 + 8 + 10 + 6 + 3 + 5 +12 = 52

(ii) People reading exactly one news paper = a + c + g = 8 + 10 + 12 = 30.

32. Sol: Given, a + b + e + f = 21

b + c + d + e = 26

d + e + f + g = 29

Also, b + e = 14, e + f = 12, e + d = 14, and e = 8

b = 6, f = 4, d = 6 Clealry e + d + f = 18 14 and 4e d f only (C) = 11

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