If n(A ∩ B′)=9, n(A′ ∩ B)=10 and n(A ∪ B)=24, then n(A × B)
Question
If
\[ n(A\cap B’)=9, \]
\[ n(A’\cap B)=10 \]
and
\[ n(A\cup B)=24, \]
then
\[ n(A\times B)= \ ? \]
Solution
We know that:
\[ A\cup B = (A\cap B’) \cup (A’\cap B) \cup (A\cap B) \]
Therefore,
\[ n(A\cup B)=n(A\cap B’)+n(A’\cap B)+n(A\cap B) \]
Substituting the given values:
\[ 24=9+10+n(A\cap B) \]
\[ 24=19+n(A\cap B) \]
\[ n(A\cap B)=5 \]
Now,
\[ n(A)=n(A\cap B’)+n(A\cap B)=9+5=14 \]
and
\[ n(B)=n(A’\cap B)+n(A\cap B)=10+5=15 \]
Therefore,
\[ n(A\times B)=n(A)\times n(B) \]
\[ =14\times15 \]
\[ =210 \]
Hence,
\[ \boxed{210} \]