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For two sets $A$ and $B$, let $A \triangle B$ denote the set of elements which belong to $A$ or $B$ but not both. If $P= \{1,2,3,4\}, Q =\{2,3,5,6\}, R=\{1,3,7,8,9\}, S =\{2,4,9,10\},$ then the number of elements in $(P\triangle Q) \triangle (R\triangle S)$ is

  1. $9$
  2. $7$
  3. $6$
  4. $8$
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Given that,

  • $ \text{T} = \{ 1,2,3,4\} $
  • $ \text{Q} = \{ 2,3,5,6 \} $
  • $ \text{R} = \{ 1,3,7,8,9\} $
  • $ \text{S} = \{ 2,4,9,10\} $

$ \boxed{ \text{A} \triangle \text{B} = ( \text{A} \cup \text{B})  – ( \text{A} \cap \text{B})} $

$ \boxed {\text{A} \triangle \text{B} = ({ \text{A} – \text{B}) \cup (\text{B} – \text{A})}} $ 

Now, $ \text{P} \triangle \text{Q} = \{ 1,2,3,4\} \triangle \{ 2,3,5,6\} $

$ \Rightarrow \text{P} \triangle \text{Q} = \{ 1, 4, 5,6 \} $

And, $ \text{R} \triangle \text{S} = \{ 1,3,7,8,9 \} \triangle \{ 2,4,9,10 \} $

$ \Rightarrow \text{R} \triangle \text{S} = \{ 1,2,3,4,7,8,10 \} $

Thus, $ ( \text{P} \triangle \text{Q}) \triangle ( \text{R} \triangle \text{S}) = \{ 1,4,5,6 \} \triangle \{ 1,2,3,4,7,8,10 \} $

$ \Rightarrow  ( \text{P} \triangle \text{Q}) \triangle ( \text{R} \triangle \text{S}) = \{ 2,3,5,6,7,8,10 \} $

$\therefore$ The number of elements in $( \text{P} \triangle \text{Q}) \triangle (\text{R} \triangle \text{S})$ is $7.$

Correct Answer $: \text{B}$

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