How To Find Cardinality Of A Set
CARDINALITY OF SETS
Cardinality of a fix is a measure out of the number of elements in the set.
For example, let A = { -2, 0, 3, 7, 9, 11, 13 }
Here, northward(A) stands for cardinality of the gear up A
And n (A) = 7
That is, there are 7 elements in the given set A.
In case, 2 or more sets are combined using operations on sets, we can find the cardinality using the formulas given below.
Formula 1 :
northward(A u B) = north(A) + n(B) - n(A n B)
If A and B are disjoint sets, due north(A northward B) = 0
And so, n(A u B) = n(A) + n(B)
Formula two :
n(A u B u C) = n(A) + n(B) + northward(C) - north(A n B) - n(B n C) - n(A n C) + due north(A n B due north C)
If A, B and C are all disjoint sets,
northward(A n B) = 0, n(B n C) = 0, n(A n C) = 0, north(A north B due north C) = 0
Then, n(A u B) = n(A) + n(B) + n(C)
Addition Theorem on Sets
Theorem 1 :
n(AuB) = n(A) + due north(B) - n(AnB)
Theorem two :
n(AuBuC) :
= due north(A) + n(B) + n(C) - n(AnB) - north(BnC) - due north(AnC) + n(AnBnC)
Explanation :
Allow us come up to know about the following terms in details.
n(AuB) = Total number of elements related to any of the two events A & B.
north(AuBuC) = Full number of elements related to any of the three events A, B & C.
north(A) = Total number of elements related to A.
n(B) = Total number of elements related to B.
n(C) = Total number of elements related to C.
For iii events A, B & C, we have
n(A) - [n(AnB) + north(AnC) - northward(AnBnC)] :
Total number of elements related to A only.
n(B) - [n(AnB) + due north(BnC) - n(AnBnC)] :
Full number of elements related to B simply.
n(C) - [n(BnC) + n(AnC) + n(AnBnC)] :
Total number of elements related to C only.
n(AnB) :
Full number of elements related to both A & B
north(AnB) - n(AnBnC) :
Total number of elements related to both (A & B) but.
n(BnC) :
Full number of elements related to both B & C
n(BnC) - n(AnBnC) :
Total number of elements related to both (B & C) only.
north(AnC) :
Full number of elements related to both A & C
due north(AnC) - n(AnBnC) :
Total number of elements related to both (A & C) only.
For two events A & B, we have
n(A) - n(AnB) :
Total number of elements related to A only.
n(B) - due north(AnB) :
Total number of elements related to B only.
Agreement Word Problem
In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play human foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and viii play all the iii games.
Let F, H and C stand for the set of students who play foot brawl, hockey and cricket respectively.
Venn diagram related to the above state of affairs :
From the venn diagram, nosotros can have the following details.
No. of students who play pes ball = 65
No. of students who play foot ball only = 28
No. of students who play hockey = 45
No. of students who play hockey merely = 18
No. of students who play cricket = 42
No. of students who play cricket just = 10
No. of students who play both pes ball & hockey = 20
No. of students who play both (human foot ball & hockey) only = 12
No. of students who play both hockey & cricket = fifteen
No. of students who play both (hockey & cricket) merely = 7
No. of students who play both pes ball and cricket = 25
No. of students who play both (foot brawl and cricket) just = 17
No. of students who play all the three games = 8
Solving Word Problem
In a group of students, 65 play human foot brawl, 45 play hockey, 42 play cricket, twenty play foot ball and hockey, 25 play foot brawl and cricket, 15 play hockey and cricket and 8 play all the 3 games. Find the full number of students in the group (Assume that each student in the grouping plays at least 1 game).
Step i :
Let F, H and C represent the set of students who play foot brawl, hockey and cricket respectively.
Step 2 :
From the given information, we have
n(F) = 65 , n(H) = 45, n(C) = 42,
north(FnH) = 20, n(FnC) = 25, n(HnC) = 15
n(FnHnC) = 8
Step 3 :
From the bones stuff, we have
Total number of students in the group is north(FuHuC).
n(FuHuC) is equal to
= due north(F) + n(H) + northward(C) - n(FnH) - n(FnC) - north(HnC) + n(FnHnC)
So, nosotros have
north(FuHuC) = 65 + 45 + 42 -20 - 25 - 15 + 8
north(FuHuC) = 100
So, the full number of students in the grouping is 100.
Culling Method (Using venn diagram) :
Step 1 :
Venn diagram related to the information given in the question :
Pace 2 :
Total number of students in the grouping :
= 28 + 12 + 18 + 7 + 10 + 17 + 8
= 100
And then, the full number of students in the group is 100.
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