For any three sets A- B - C-nA-a- nB-b - nC-c- then nB- C- - - nA- B- C-

Given three sets: A, B & C such that nbsp; n(A)=a, n(B)=b & n(C)=c, Thus, n(B× C)=n(B)× n(C)=bc And similarly, nbsp; n(A× B× C)=n(A)× n(B)× n(C)=ab ...

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Standard XII

Mathematics

Number of Relations for Two or More Sets

For any three...

Question

For any three sets $A, B & C,$
$n (A) = a, n (B) = b & n (C) = c,$ then $n (B \times C) =$ , $& n (A \times B \times C) =$ .

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Solution

Given three sets: $A, B & C$ such that
$n (A) = a, n (B) = b & n (C) = c,$
Thus, $n (B \times C) = n (B) \times n (C) = b c$
And similarly,
$n (A \times B \times C) = n (A) \times n (B) \times n (C) = a b c$

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Similar questions

Q. Observe the figure 1.18 and verify the following equations.
n (A

\cup

\cup

C) = n (A) + n (B) + n (C) − n (A

\cap

B ) − n (B

\cap

C) − n (C

\cap

A) + n (A

\cap

\cap

C)

figure

Q. For any three sets

A, B & C,

n (A \cup B \cup C) = n (A) + n (B) + n (C)

- n (A \cap B) - n (B \cap C) - n (C \cap A)

+ n (A \cap B \cap C)

Q. Verify n(A

\cup

\cup

C) = n(A) + n(B) + n(C) – n(A

\cap

B) – n(B

\cap

C) – n(A

\cap

C) + n(A

\cap

\cup

C) for the following sets. A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}

Q. Let

A, B, C

are three sets such that

n (A \cap B) = n (B \cap C) = n (C \cap A) = n (A \cap B \cap C) = 2

, then

n ((A \times B) \cap (B \times C))

is equal to

Q. Verify

n (A \cup B \cup C) = n (A) + n (B) + n (c) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)

for the sets given below:
(i) A={4,5,6}, B={5,6,7,8} and C={6,7,8,9}
(ii) A={a, b, c, d, e}, B={x, y, z} and C={a, e, x}

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For any three sets A,B & C, n(A)=a,n(B)=b & n(C)=c, then n(B×C)= , & n(A×B×C)=.

Given three sets: A,B & C such that n(A)=a,n(B)=b & n(C)=c, Thus, n(B×C)=n(B)×n(C)=bc And similarly, n(A×B×C)=n(A)×n(B)×n(C)=abc

For any three sets $A, B & C,$
$n (A) = a, n (B) = b & n (C) = c,$ then $n (B \times C) =$ , $& n (A \times B \times C) =$ .

Given three sets: $A, B & C$ such that
$n (A) = a, n (B) = b & n (C) = c,$
Thus, $n (B \times C) = n (B) \times n (C) = b c$
And similarly,
$n (A \times B \times C) = n (A) \times n (B) \times n (C) = a b c$