WBJEE · Maths · Sets and Relations
Let \(S=\{(a, b, c) \in N \times N \times N: a+b+c=21\), \(a \leq b \leq c\}\) and \(T=\{(a, b, c) \in N \times N \times N: a, b, c\) are in \(A P \}\) where \(N\) is the set of all natural numbers. Then, the number of elements in the set \(\mathrm{S} \cap T\)
- A 6
- B 7
- C 13
- D 14
Answer & Solution
Correct Answer
(B) 7
Step-by-step Solution
Detailed explanation
We have, \(a+b+c=21\) and \(\frac{a+c}{2}=b\) \(\Rightarrow \quad a+c+\frac{a+c}{2}=21\) \(\begin{array}{ll}\Rightarrow & a+c=14 \Rightarrow \frac{a+c}{2}=7 \\ \Rightarrow & b=7\end{array}\) So, a can take values from 1 to 6 . C can have values from 8 to 13 or a=b=c=7 So, there…
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