Plus One Maths | Chapter 1 | Sets | Full Chapter | Exam Winner

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Embark on a journey through CHAPTER 1 - SETS in our Plus One Maths tutorial. In this comprehensive lesson, we'll unravel the fundamental principles and applications of sets.
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The given information is as follows:
(n(A) = 10)
(n(B) = 17)
(n(A U B) = 23)
Substituting the given values: [ n(AU B) = 10 + 17 - 23 = 4 ]
Therefore, there are 4 elements common to both sets A and B. 😊

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12:30

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Ans: 10+17-23=4
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Ans:4

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Ans 4

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1:07:14

[@Rajupallakkad]

Ans)Formula = n(AUB) = n(A)+ n(B)-n(A^B)
n(A^B) = n(A)+n(B)-n(AUB)
= 10+17-23
= 27-23
= 4
Therefore , n(A^B)= 4
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Question:
n(A)=10
n(B)=17
n(AUB)=23
n(A^B)=?
Solution:
n(AUB)= n(A)+n(B)-n(A^B)
23= 10+17-n(A^B)
23= 27-n(A^B)
n(A^B)= 27-23
n(A^B)= 4
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Ans:👇
n(A)=10
n(B)=17
n(AUB)=23
n(AUB)=n(A)+n(B)-n(A^B)
23=10+17-n(A^B)
23=27-n(A^B)
27-23=n(A^B)
4=n(A^B)

[@pavithrapj5020]

10+17-23
=27-23
=4

[@Sheejasheejaa]

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[@sreejavr5022]

n(AUB)=n(A)+n(B)-n(A^B)
n(A^B)=n(A)+n(B)-n(AUB)
=10+17-23
=4

[@soulofdream5678]

Ans.10+17 - 23 = 4
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[@Achukittycat]

Ans 4
Thank you sir..

[@shamlashinaj]

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Answer:4

[@mayukhmv8033]

n(AnB)=4

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answer=A intersection B=4

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[@ABHI__444]

N(A) +n(B) -n(AUB)
10+17-23= 4

[@PriyaBiju-sf8er]

n(Anb)=n(A)+n(B)-n(AUB)
=10+17-50
=4

[@Anand-ds7pw]

Ans:4
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[@AJO....777]

23=10+17-n(A n B)
(A n B)=4
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Ans = 4

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10+17-23=4

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Ans : 4

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21:49

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[@boltpc6283]

Ans. 4

[@shamlashinaj]

n(AnB)=4

[@BiggBossmalayalamliveseason5]

1:40:23

[@shafeekmalayady1894]

Ans 4